## Tag Archives: group

### Equivalent characterizations of groups

Let $S$ be a semigroup. Prove that the following are equivalent.

1. For every $a \in S$, there exists a unique $x \in S$ such that $ax$ is idempotent.
2. For every $a \in S$, there exists a unique element $x \in S$ such that $a = axa$.
3. $S$ is regular and contains exactly one idempotent.
4. $S$ is a group.

We will follow the strategy $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4)$.

$(1) \Rightarrow (2)$: Suppose that for every $a \in S$, there exists a unique $x \in S$ such that $ax$ is idempotent. Let $a \in S$. Now there exists a unique $x \in S$ such that $axax = ax$. Note that $a(xax)a(xax) = a(xax)$, so that (by the uniqueness of $x$) we have $x = xax$. Now $xaxa = xa$ and $x(axa)x(axa) = x(axa)$, so that $a = xax$. Suppose now that $a = aya$ for some $y \in S$. Then $ayay = ay$, and again by uniqueness, $y = x$. So for all $a \in S$ there exists a unique $x \in S$ such that $a = axa$.

$(2) \Rightarrow (3)$: Suppose that for all $a \in S$ there exists a unique $x \in S$ with $a = axa$. Now $S$ is regular by definition. Now $a(xax)a = axa = a$, so that by uniqueness, $x = xax$. So $x$ is an inverse for $a$. Now let $e \in S$ be idempotent (these exist in $S$ since for example $ax$ is idempotent) and let $a \in S$ with inverse $x$. Now $ea = eayea$ for some $y \in S$, so that $y = yeay$. Now $ye = yeaye$, so that $a = ayea$. So $ye = x$. Now $xe = ye^2 = ye = x$. Since every element of $S$ is the inverse of some other element, in fact $e$ is a right identity. A similar argument shows that $e$ is a left identity. So every idempotent is an identity. Since a semigroup can have at most one identity, $S$ has a unique idempotent.

$(3) \Rightarrow (4)$: Suppose $S$ is regular and has a unique idempotent $e$. Now for any $a \in S$, there exists an element $x \in S$ such that $a = axa$. Now $axax = ax$ and $xaxa = xa$, so that $xa$ and $ax$ are idempotent. Thus $ax = xa = e$. In particular, we have $ea = ae = a$ for all $a \in S$. So $e \in S$ is an identity element, and every $a \in S$ has an inverse with respect to $e$. So $S$ is a group.

$(4) \Rightarrow (1)$: Suppose $S$ is a group with identity $e$. Denote the inverse of $a$ by $a^{-1}$. Certainly $aa^{-1} = e$ is idempotent. Now if $x \in S$ is idempotent, we have $xx = x$, so that $xx^{-1} = xx^{-1}$, so $x = e$. So if $ab = e$ is idempotent, then $a^{-1}ab = a^{-1}$ and thus $b = a^{-1}$. That is, for every $a \in S$ there is a unique $x \in S$ such that $ax$ is idempotent.

### Groups are precisely those semigroups which are both left and right simple

Prove that a semigroup $S$ is a group if and only if it is both left and right simple. Exhibit a left simple semigroup which is not right simple.

We begin with a lemma.

Lemma: Let $S$ be a semigroup. Then $S$ is left simple (right simple) [simple] if and only if $Sa = S$ ($aS = S$) [$SaS = S$] for all $a \in S$. Proof: Suppose $S$ is left simple, and let $a \in S$. Certainly $SSa \subseteq Sa$, so that $Sa$ is a left ideal. Thus $Sa = S$ for all $a \in S$. Conversely, suppose $Sa = S$ for all $a \in S$, and let $L \subseteq S$ be a left ideal. Now $L$ is nonempty by definition; say $a \in L$. Then $Sa \subseteq L$, and so $S \subseteq L$. Thus $L = S$, and in fact $S$ is the only left ideal of $S$. So $S$ is left simple. The proofs for right simple and simple are similar. $\square$

Now let $S$ be a group and let $a \in S$. If $x \in S$, then we have $x = xe = xa^{-1}a$; in particular, $x \in Sa$. So $S = Sa$ for all $a$. By the lemma, $S$ is left simple. Similarly, $S$ is right simple.

Now suppose the semigroup $S$ is both left and right simple. Let $a \in S$. Since $Sa = S$, there exists an element $e \in S$ such that $ea = a$. Now let $b \in S$ be arbitrary. Since $aS = S$, there exists $c \in S$ such that $b = ac$. Now $b = ac = eac = eb$, so in fact $e$ is a left identity element of $S$. Similarly, there is a right identity element $f$, and we have $e = ef = f$, so that $e$ is a two-sided identity.

Now let $a \in S$ be arbitrary. Since $Sa = aS = S$, there exist elements $b,c \in S$ such that $ba = ac = e$. Now $b = be = bac = ec = c$, and we have $b = c$. Thus $a$ has a two sided inverse with respect to $e$. Since $a$ was arbitrary, every element has an inverse, and so $S$ is a group.

Now consider the semigroup $S = \{a,b\}$ with $xy = x$ for all $x,y \in S$. (That is, $S$ is the left zero semigroup on two elements.) Suppose $T \subseteq S$ is a left ideal. Now for all $x \in S$ and $y \in T$, we have $xy = x \in T$. Thus $T = S$, and so $S$ is left simple. However, $S$ is not a group, and so (by the previous argument) cannot be right simple.

### Some properties of a commutative diagram of groups with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact. That is, $\mathsf{ker}\ \varphi_k = \mathsf{im}\ \psi_k$ and $\mathsf{ker}\ \chi_k = \mathsf{im}\ \varphi_k$ for $k \in \{1,2\}$. Prove the following.

1. If $\alpha$ is surjective and $\beta$ and $\delta$ are injective, then $\gamma$ is injective.
2. If $\delta$ is injective and $\alpha$ and $\gamma$ are surjective, then $\beta$ is surjective.

1. Let $c \in \mathsf{ker}\ \gamma$. Note that $\chi_2(\gamma(c)) = 1$, so that $1 = (\chi_2 \circ \gamma)(c) = (\delta \circ \chi_1)(c) = \delta(\chi_1(c))$. Since $\delta$ is injective, $\chi_1(c) = 1$. Thus $c \in \mathsf{ker}\ \chi_1$, and so $c \in \mathsf{im}\ \varphi_1$. Say $b \in B_1$ such that $c = \varphi_1(b)$. Now $1 = \gamma(c)= \gamma(\varphi_1(b))$ $= (\gamma \circ \varphi_1)(b)$ $= (\varphi_2 \circ \beta)(b)$ $= \varphi_2(\beta(b))$, so that $\beta(b) \in \mathsf{ker}\ \varphi_2$. Thus $\beta(b) \in \mathsf{im}\ \psi_2$. Say $a^\prime \in A_2$ with $\beta(b) = \psi_2(a^\prime)$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $\beta(b) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a)$ $= \beta(\psi_1(a))$. Since $\beta$ is injective, we have $b = \psi_1(a)$. In particular, $b \in \mathsf{im}\ \psi_1$, so that $b \in \mathsf{ker}\ \varphi_1$. Thus $c = \varphi_1(b) = 1$. So $\mathsf{ker}\ \gamma = 1$, and so $\gamma$ is injective.
2. Let $b \in B_2$. Now $\varphi_2(b) \in C_2$. Since $\gamma$ is surjective, there exists $c^\prime \in C_1$ such that $\gamma(c^\prime) = \varphi_2(b)$. Now $\gamma(c^\prime) \in \mathsf{im}\ \varphi_2$, so that $\gamma(c^\prime) \in \mathsf{ker}\ \chi_2$. Now $(\chi_2 \circ \gamma)(c^\prime) = 1$, so that $(\delta \circ \chi_1)(c^\prime) = 1$. Since $\delta$ is injective, $\chi_1(c^\prime) = 1$. Thus $c^\prime \in \mathsf{ker}\ \chi_1$. So $c^\prime \in \mathsf{im}\ \varphi_1$. Say $b^\prime \in B_1$ such that $\varphi_1(b^\prime) = c^\prime$. Now $\varphi_2(b) = \gamma(c^\prime)$ $= (\gamma \circ \varphi_1)(b^\prime)$ $= (\varphi_2 \circ \beta)(b^\prime)$ $= \varphi_1(\beta(b^\prime))$. Thus $\varphi_2(b^{-1}\beta(b^\prime)) = 1$, so that $b^{-1}\beta(b^\prime) \in \mathsf{ker}\ \varphi_2$. Now $b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2$; say $a \in A_2$ such that $b^{-1}\beta(b^\prime) = \psi_2(a)$. Since $\alpha$ is surjective, there exists $a^\prime \in A_1$ such that $\alpha(a^\prime) = a$. Now $b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a^\prime)$ $= (\beta \circ \psi_1)(a^\prime)$. Thus $b^{-1}\beta(b^\prime) = \beta(\psi_1(a^\prime))$, and so $b = \beta(b^\prime \psi_1((a^\prime)^{-1}))$. Thus $\beta$ is surjective.

### Some properties of a commutative diagram with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact; that is, $\mathsf{im}\ \psi_k = \mathsf{ker}\ \varphi_k$ for $k \in \{1,2\}$. Prove the following.

1. If $\varphi_1$ and $\alpha$ are surjective and $\beta$ is injective, then $\gamma$ is injective.
2. If $\psi_2$, $\alpha$, and $\gamma$ are injective, then $\beta$ is injective.
3. If $\varphi_1$, $\alpha$, and $\gamma$ are surjective, then $\beta$ is surjective.
4. If $\beta$ is surjective and $\gamma$ and $\psi_2$ are injective, then $\alpha$ is surjective.

1. Let $c \in \mathsf{ker}\ \gamma$. Since $\varphi_1$ is surjective, there exists $b \in B_1$ such that $\varphi(b) = c$. Now $(\gamma \circ \varphi_1)(b) = 1$, so that $(\varphi_2 \circ \beta)(b) = 1$, and so $\varphi_2(\beta(b)) = 1$. Thus $\beta(b) \in \mathsf{ker}\ \varphi_2$, and so $\beta(b) \in \mathsf{im}\ \psi_2$. Say $\beta(b) = \psi_2(a^\prime)$ where $a^\prime \in A_2$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $\beta(b) = (\psi_2 \circ \alpha)(a)$, so that $\beta(b) = (\beta \circ \psi_1)(a)$, and so $\beta(b) = \beta(\psi_1(a))$. Since $\beta$ is injective, $b = \psi_1(a)$. In particular, $b \in \mathsf{im}\ \psi_1$, so that $b \in \mathsf{ker}\ \varphi_1$. Then $c = \varphi_1(b) = 1$. Thus $\mathsf{ker}\ \gamma = 1$, and so $\gamma$ is injective.
2. Let $b \in \mathsf{ker}\ \beta$. Now $\beta(b) = 1$, so that $(\varphi_2 \circ \beta)(b) = 1$, so that $(\gamma \circ \varphi_1)(b) = 1$. Thus $\gamma(\varphi_1(b)) = 1$. Since $\gamma$ is injective, $\varphi_1(b) = 1$, so that $b \in \mathsf{ker}\ \varphi_1$. So $b \in \mathsf{im}\ \psi_1$. Thus there exists $a \in A_1$ such that $b = \psi_1(a)$. Now $(\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a))$ $= \beta(b) = 1$, so that $a \in \mathsf{ker}\ \psi_2 \circ \alpha$. Since $\alpha$ and $\psi_2$ are injective, $\psi_2 \circ \alpha$ is injective, so that $a = 1$. Thus $b = \psi_1(1) = 1$, and thus $\mathsf{ker}\ \beta = 1$. Thus $\beta$ is injective.
3. Let $b \in B_2$. Now $\varphi_2(b) \in C_2$. Since $\gamma$ and $\varphi_1$ are surjective, $\gamma \circ \varphi_1$ is surjective, so that there exists $b^\prime \in B_1$ such that $\varphi_2(b) = (\gamma \circ \varphi_1)(b^\prime)$. Note that $\varphi_2(b^{-1}\beta(b^\prime)) = 1$, so that $b^{-1} \beta(b^\prime) \in \mathsf{ker}\ \varphi_2$. Thus $b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2$, and so there exists $a^\prime \in A_2$ such that $b^{-1}\beta(b^\prime) = \psi_2(a^\prime)$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a))$. Thus $b = \beta(b^\prime \psi_1(a^{-1}))$, and so $\beta$ is surjective.
4. Let $a \in A_2$. Now $\psi_2(a) \in B_2$. Since $\beta$ is surjective, there exists $b^\prime \in B_1$ such that $\psi_2(a) = \beta(b^\prime)$. Now $1 = (\varphi_2 \circ \psi_2)(a) = \varphi_2(\psi_2(a))$ $= \varphi_2(\beta(b^\prime))$ $= (\varphi_2 \circ \beta)(b^\prime)$ $= (\gamma \circ \varphi_1)(b^\prime)$ $= \gamma(\varphi_1(b^\prime))$. Since $\gamma$ is injective, $\varphi_1(b^\prime) = 1$. So $b^\prime \in \mathsf{ker}\ \varphi_1$, and so $b^\prime \in \mathsf{im}\ \psi_1$. Say $a^\prime \in A_1$ such that $b^\prime = \psi_1(a^\prime)$. Now $\psi_2(a) = \beta(b^\prime)$ $= \beta(\psi_1(a^\prime))$ $= (\beta \circ \psi_1)(a^\prime)$ $= (\psi_2 \circ \alpha)(a^\prime)$ $= \psi_2(\alpha(a^\prime))$. Since $\psi_2$ is injective, $a = \alpha(a^\prime)$. Thus $\alpha$ is surjective.

### Count the number of automorphisms of a group from a presentation

Use presentations to find the orders of the automorphism groups of $Z_2 \times Z_4$ and $Z_4 \times Z_4$.

We have $Z_2 \times Z_4 = \langle a,b \ |\ a^2 = b^4 = 1, ab = ba \rangle$.

Now any element $x$ of order 4 and any element $y$ of order 2 with $x \neq y^2$ generate $Z_2 \times Z_4$. There are 4 elements of order 4: $b$, $b^3$, $ab$, and $ab^3$. There are 2 elements of order 2: $a$, $b^2$, and $ab^2$. Once $x$ is chosen, there are two choices for $y$, and each choice determines an automorphism. These are distinct by construction. Thus $|\mathsf{Aut}(Z_2 \times Z_4)| = 8$.

Now $Z_4 \times Z_4 = \langle a,b \ |\ a^4 = b^4 = 1, ab = ba \rangle$.

Any two elements of order 4, say $x$ and $y$, generate $Z_4 \times Z_4$ provided $\langle x \rangle \cap \langle y \rangle = 1$. This group has 12 elements of order 4, and these intersect nontrivially pairwise. Thus once $x$ is chosen, there are 8 choices for $y$. Thus $|\mathsf{Aut}(Z_4 \times Z_4)| = 96$.

### A group of order 24 with no elements of order 6 is isomorphic to Sym(4)

Show that a group of order 24 with no element of order 6 is isomorphic to $S_4$.

Note that $24 = 2^3 \cdot 3$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) = 1$. If $n_2(G) = 1$, then by the recognition theorem for direct products, $G \cong P_2 \times P_3$, where $P_2$ and $P_3$ are Sylow 2- and 3-subgroups of $G$, respectively. By Cauchy, there exist elements $x \in P_2$ and $y \in P_3$ of order 2 and 3, so that $xy$ has order 6, a contradiction. Suppose now that $n_2(G) = 3$. Since $n_2(G) \not\equiv 1$ mod 4, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $P_2 \cap Q_2$ is nontrivial and its normalizer has order $2^3 \cdot 3$, by a previous theorem. Thus $P_2 \cap Q_2 \leq G$ is normal. Note that $|P_2 \cap Q_2|$ is either 2 or 4.

Suppose $|P_2 \cap Q_2| = 4$. Then at most $7 + 3 + 7$ nonidentity elements of $G$ are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that $|P_2 \cap Q_2| = 2$. By the N/C Theorem, $G/C_G(P_2 \cap Q_2) \leq \mathsf{Aut}(P_2 \cap Q_2) \cong 1$, so that $P_2 \cap Q_2$ is central in $G$. By Cauchy, there exist elements $x \in P_2 \cap Q_2$ and $y \in G$ of order 2 and 3, so that $xy$ has order 6, a contradiction.

Thus we may assume $n_3(G) = 4$. Let $P_3 \leq G$ be a Sylow 3-subgroup and let $N = N_G(P_3)$. The action of $G$ on $G/N$ yields a permutation representation $G \rightarrow S_4$ whose kernel $K$ is contained in $N$. Recall that normalizers of Sylow subgroups are self normalizing, so that $N$ is not normal in $G$. Moreover, we have $|N| = 6$. We know from the classification of groups of order 6 that $N$ is isomorphic to either $Z_6$ or $D_6$; however, in the first case we have an element of order 6, a contradiction. Thus $N \cong D_6$. We know also that the normal subgroups of $D_6$ have order 1, 3, or 6. If $|K| = 6$, then $K = N$ is normal in $G$, a contradiction. If $|K| = 3$, then by the N/C theorem we have $G/C_G(K) \leq \mathsf{Aut}(Z_3) \cong Z_2$. In particular, $C_G(K)$ contains an element of order 2, so that $G$ contains an element of order 6, a contradiction.

Thus $K = 1$, and in fact $G \leq S_4$. Since $|G| = |S_4| = 24$ is finite, $G \cong S_4$.

### The Frattini subgroup is the set of all nongenerators

An element $x$ of a group $G$ is called a nongenerator if for all proper subgroups $H \leq G$, $\langle x, H \rangle \leq G$ is also proper. Prove that $\Phi(G) \leq G$ is the set $\mathsf{ng}(G)$ of all nongenerators in $G$.

$(\subseteq)$ [Not finished.] Let $x \in \Phi(G)$. Now let $H < G$ be a proper subgroup. If $H$ is contained in some maximal subgroup $M$, then we have $\Phi(G)H \leq M < G$, so that $\langle x, H \rangle < G$ is proper as desired. Suppose now that $H$ is not contained in any maximal subgroup of $G$. Suppose further that $\Phi(G)H = G$. (@@@)

[Old proof, assumes $G$ is finite.]
Let $x \in \Phi(G)$. By a lemma we proved for a previous theorem, if $H \leq G$ is proper then $\Phi(G)H \leq G$ is proper. Now $\langle x \rangle \leq \Phi(G)H$ and $H \leq \Phi(G)H$, so that $\langle x,H \rangle \leq \Phi(G)H < G$ is proper. Thus $x \in \mathsf{ng}(G)$ is a nongenerator.

$(\supseteq)$ Let $x \in \mathsf{ng}(G)$ be a nongenerator of $G$. If $G$ has no maximal subgroups, then $\Phi(G) = G$, and we have $\mathsf{ng}(G) \subseteq \Phi(G)$. Suppose now that $M \leq G$ is maximal; in particular, $M \leq G$ is proper, so that $\langle x, M \rangle \leq G$ is proper. Thus $\langle x, M \rangle \leq M$, so that $x \in M$. Since $x$ is contained in every maximal subgroup, $x \in \Phi(G)$.

### Thirteen nonisomorphic groups of order 56

This exercise describes thirteen isomorphism types of groups of order 56. (It is not too difficult to show that every group of order 56 is isomorphic to one of these.)

1. Prove that there are three abelian groups of order 56.
2. Prove that every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.
3. Construct the following nonabelian groups of order 56 which have a notmal Sylow 7-subgroup and whose Sylow 2-subgroup $K$ is as specified.
1. One group with $K \cong Z_2^3$
2. Two nonisomorphic groups when $K \cong Z_4 \times Z_2$
3. One group with $K \cong Z_8$
4. Two nonisomorphic groups with $K \cong Q_8$
5. Three nonisomorphic groups with $K \cong D_8$
4. Let $G$ be a group of order 56 with a nonnormal Sylow 7-subgroup. Prove that if $K$ is the Sylow 2-subgroup of $G$ then $S \cong Z_2^3$.
5. Prove that there is a unique group of order 56 with a nonnormal Sylow 7-subgroup.

We begin with a lemma.

Lemma 1: Let $H$ and $K$ be groups and $\varphi : K \rightarrow \mathsf{Aut}(H)$ a group homomorphism. If $H \rtimes_\varphi K$ is abelian, then $\varphi$ is trivial. Proof: We may assume that $H$ and $K$ are abelian, since both are subgroups of $H \rtimes_\varphi K$. Let $h \in H$ and $k \in K$. Note that $(h,1)(h,k) = (h \varphi(1)(h), k)$ and $(h,k)(h,1) = (h \varphi(k)(h), k)$. Since $H \rtimes_\varphi K$ is abelian, these are equal. Comparing entries we have $h \varphi(1)(h) = h \varphi(k)(h)$, so that $\varphi(k)(h) = h$ for all $h$. Thus $\varphi(k) = 1$ for all $k$; hence $\varphi$ is trivial.

1. Note that $56 = 2^3 \cdot 7$. By FTFGAG, the nonisomorphic abelian groups of order 56 are $Z_{56}$, $Z_{28} \times Z_2$, and $Z_{14} \times Z_2 \times Z_2$.
2. Let $G$ be a group of order 56. By Sylow’s Theorem, $n_2 \in \{ 1,7 \}$ and $n_7 \in \{1,8\}$. Note that the Sylow 7-subgroups of $G$ intersect trivially. If $n_2 = 7$ and $n_7 = 8$, then $G$ has $6 \cdot 8 = 48$ elements of order 7, 7 nonidentity elements in some Sylow 2-subgroup $P$, 1 element in some Sylow 2-subgroup $Q$ which is not in $P$, and an identity, for a total of at least 57 elements- a contradiction. Thus at least one of $n_2$ and $n_7$ is 1.
3. Let $H = Z_7 = \langle x \rangle$. Now $\mathsf{Aut}(H) \cong Z_6$; say $\mathsf{Aut}(H) = \langle \alpha \rangle$, where $\alpha(x) = x^2$. Note that $\mathsf{Aut}(H)$ has a unique subgroup of order dividing 8; namely $\langle \alpha^3 \rangle$.
1. Let $K = Z_2^3 = \langle a \rangle \times \langle b \rangle \times \langle c \rangle$. By a previous theorem, there is a unique homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$ such that $\varphi(a) = \varphi(b) = \varphi(c) = \alpha^3$. Now $Z_7 \rtimes_\varphi Z_2^3$ is a group of order 56 which is nonabelian by Lemma 1.
2. Let $K = Z_4 \times Z_2 = \langle a \rangle \times \langle b \rangle$. By a previous theorem, there exist unique homomorphisms $\varphi, \psi : K \rightarrow \mathsf{Aut}(H)$ such that $\varphi(a) = \alpha^3$ and $\varphi(b) = 1$ while $\psi(a) = 1$ and $\psi(b) = \alpha^3$. We can see that $\mathsf{ker}\ \varphi = \langle a^2 \rangle \times \langle b \rangle \cong Z_2 \times Z_2$ while $\mathsf{ker}\ \psi = \langle a \rangle \times 1 \cong Z_4$; in particular these kernels are not isomorphic. Thus $Z_7 \rtimes_\varphi (Z_4 \times Z_2)$ and $Z_7 \rtimes_\psi (Z_4 \times Z_2)$ are nonisomorphic groups of order 56, and are nonabelian by Lemma 1.
3. Let $K = Z_8 = \langle a \rangle$. There is a unique nontrivial group homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$ given by $\varphi(a) = \alpha^3$. Using Lemma 1, $Z_7 \rtimes_\varphi Z_8$ is a nonabelian group of order 56.
4. Let $K = Q_8 = \langle i,j \rangle$. Note that every group homomorphism $\varphi : Q_8 \rightarrow Z_6$ is determined by its values at $i$ and $j$. Moreover, since $|\varphi(i)|$ and $|\varphi(j)|$ must divide 4, we have $\varphi(i), \varphi(j) \in \{1, \alpha^3 \}$. Now if $\varphi(i) = \varphi(j) = 1$, then $\varphi = 1$ and $\varphi$ is in fact a homomorphism. This gives rise to the direct product $Z_7 \times Q_8$. Now suppose $\varphi(i) = \alpha^3$ and $\varphi(j) = 1$. Since $\varphi(i) \varphi(j) = \varphi(j)^3 \varphi(i)$, $\varphi$ extends to a homomorphism. Now $Z_7 \rtimes_\varphi Q_8$ is distinct from $Z_7 \times Q_8$ since $\varphi$ is nontrivial.
5. Let $K = D_8 = \langle r,s \rangle$. Note that every group homomorphism $\varphi : D_8 \rightarrow Z_6$ is determined by its values at $r$ and $s$. Moreover, since $|\varphi(r)|$ must divide 4 and $|\varphi(s)|$ must divide 2, we have $\varphi(r), \varphi(s) \in \{1, \alpha^3 \}$.
1. If $\varphi(r) = \varphi(s) = 1$, then $\varphi = 1$ is in fact a homomorphism. Then $Z_7 \rtimes_\varphi D_8 \cong Z_7 \times D_8$.
2. If $\varphi(r) = \alpha^3$ and $\varphi(s) = 1$, then since $\varphi(r) \varphi(s) = \varphi(s)\varphi(r)^3$, $\varphi = \varphi_1$ is in fact a homomorphism. We can see that $\mathsf{ker}\ \varphi_1 = \langle r^2, s \rangle \cong Z_2^2$. Since $\varphi_1$ is nontrivial, $Z_7 \rtimes_{\varphi_1} D_8$ is a nonabelian group of order 56.
3. If $\varphi(r) = 1$ and $\varphi(s) = \alpha^3$, then since $\varphi(r) \varphi(s) = \varphi(s) \varphi(r)^3$, $\varphi = \varphi_2$ is in fact a homomorphism. We can see that $\mathsf{ker}\ \varphi_2 = \langle r \rangle \cong Z_4$. Since $\varphi_2$ is nontrivial, $Z_7 \rtimes_{\varphi_2} D_8$ is a nonabelian group of order 56.

Now these three groups are distinct since $\varphi_1$ and $\varphi_2$ are nontrivial and $\mathsf{ker}\ \varphi_1 \not\cong \mathsf{ker}\ \varphi_2$.

4. Let $G$ be a group of order 56 and with Sylow 2- and 7-subgroups $P_2$ and $P_7$, respectively. Suppose $P_7$ is not normal; then in particular, $G$ is not abelian. By part (2) above, $P_2$ is normal. By Lagrange, $P_2 \cap P_7 = 1$, and in fact $G$ is the internal semidirect product $P_2 \rtimes_\varphi P_7$ by the recognition theorem for semidirect products, where $\varphi(x)(y) = xyx^{-1}$. Thus $P_7$ acts on $P_2$ by conjugation. More strongly, $P_7$ acts on the set $X$ of nonidentity elements of $P_2$, since the identity is in its own conjugacy class.

We claim that this action is transitive. To see why, note that by the Orbit-Stabilizer Theorem, the orbits of this action have cardinality 1 or 7. Moreover, if some orbit has cardinality 1, then all orbits have cardinality 1. However, in this case the mapping $\varphi$ is trivial and we have $G \cong P_7 \times P_2$, contradicting the nonnormalcy of $P_7$ in $G$. Thus some – hence the only – orbit of this action has cardinality 7 and the action is transitive. In particular, all nonidentity elements of $P_2$ have the same order. By Cauchy’s Theorem, some nonidentity element has order 2; thus all nonidentity elements of $P_2$ have order 2. So $P_2$ is elementary abelian of order $8$, that is, $Z_2^3$.

5. (Existence) Let $H = Z_2^3$ and $K = Z_7 = \langle x \rangle$. We know that $\mathsf{Aut}(Z_2^3) \cong GL_3(\mathbb{F}_2)$. Since $|GL_3(\mathbb{F}_2)| = 168 = 7 \cdot 24$, by Cauchy’s Theorem there exists an element $\alpha \in \mathsf{Aut}(H)$ of order 7. By a previous theorem there exists a unique group homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$ such that $\varphi(x) = \alpha$. Now $H \rtimes_\varphi K$ is a group of order 56 and $K$ a Sylow 7-subgroup; moreover, by Theorem 11 in the text, $K$ is not normal since $\varphi$ is not trivial.

(Uniqueness) Let $G$ be a group of order 56 with a nonnormal Sylow 7-subgroup $K \cong Z_7$. By part (4) above, the (normal) Sylow 2-subgroup $H$ of $G$ is isomorphic to $Z_2^3$. By the recognition theorem for semidirect products, $G = H \rtimes_\psi K$ for some group homomorphism $\psi : K \rightarrow \mathsf{Aut}(H)$. Since $K \leq G$ is not normal, $\psi$ is not trivial. Thus $\mathsf{ker}\ \psi = 1$ (since $K$ is simple) so that $\psi$ is injective. Since $|\mathsf{Aut}(H)| = 168 = 7 \cdot 24$, $\mathsf{im}\ \psi$ is a Sylow 7-subgroup of $\mathsf{Aut}(H)$. By Sylow’s Theorem, $K$ is conjugate to the subgroup $\langle \alpha \rangle$ constructed above. Since $K$ is cyclic, by a previous theorem we have $H \rtimes_\psi K \cong H \rtimes_\varphi K$.

Thus there exists a unique group of order 56 with a nonnormal Sylow 7-subgroup.

### A fact regarding the interaction of powers and commutators in a group

Let $G$ be a group and let $x,y \in G$ such that $x$ and $y$ commute with $[x,y]$. Prove that $(xy)^n = x^ny^n[y,x]^{n(n-1)/2}$.

We begin with some lemmas.

Lemma 1: Let $a,b \in G$ be group elements and $n$ a positive integer. Then $(ab)^n = a(ba)^{n-1}b$. Proof: We proceed by induction on $n$. If $n = 1$, the conclusion is trivial. Suppose the conclusion holds for some $n$. Then $(ab)^{n+1} = (ab)^n ab$ $= a(ba)^{n-1}bab$ $= a(ba)^nb$, and the conclusion holds for all $n$. $\square$

Lemma 2: Let $x$ and $y$ be as in the problem statement and let $n$ be a positive integer. For all $1 \leq k \leq n$, we have $(xy)^n = x^k [y,x]^{k(k-1)/2} (yx[y,x]^{k-1})^{n-k} y^k$. Proof: We proceed by induction on $k$. For the base case $k = 1$, the conclusion is trivial (using Lemma 1). For the inductive step, suppose the conclusion holds for some $1 \leq k < n$. Then, noting that $yx = xy[y,x]$, we have the following.

 $(xy)^n$ $=$ $x^k [y,x]^{k(k-1)/2} (yx[y,x]^{k-1})^{n-k} y^k$ $=$ $x^k [y,x]^{k(k-1)/2} (xy[y,x]^k)^{n-k} y^k$ $=$ $x^k [y,x]^{k(k-1)/2} (x[y,x]^ky)^{n-k} y^k$ $=$ $x^k [y,x]^{k(k-1)/2} x[y,x]^k (yx[y,x]^k)^{n-k-1} yy^k$ $=$ $x^{k+1} [y,x]^{(k+1)k/2} (yx[y,x]^k)^{n-(k+1)} y^{k+1}$

Thus the conclusion holds for all $1 \leq k \leq n$. $\square$

Now the main result follows from Lemma 2 with $k=n$.

### Exhibit an infinite group with finite exponent

Prove that any finite group has finite exponent. Give an example of an infinite group with finite exponent. Does a finite group of exponent $m$ always contain an element of order $m$?

Let $G$ be a finite group of order $m$. By Lagrange, $x^m = 1$ for all $x \in G$, so that the exponent of $G$ is finite and divides $m$.

Consider the infinite group $G = \prod_{\mathbb{N}} Z_2$. We saw previously (indeed it is obvious) that every element of this group has order 1 or 2, so that $G$ has exponent 2.

Consider the finite group $D_6$. The exponent of $D_6$ divides 6, and is not 1 (since $D_6$ is not trivial), 2 (since $r^2 \neq 1$), or 3 (since $s^3 = s \neq 1$). Thus the exponent of $D_6$ is 6; however $D_6$ contains no element of order 6 because (for one) it is not abelian, hence not cyclic.