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.