Tag Archives: subgroup

Every finite multiplicative subgroup of a field is cyclic

Let $F$ be a field and $G \leq F^\times$ be a finite multiplicative subgroup. Prove that $G$ is cyclic.

We prove this first in the case that $G$ is a $p$-group, say $|G| = p^k$, by induction on $k$. Given $G$, we let $\psi(d)$ denote the number of elements of $G$ of order $d$.

Suppose $|G| = p$; then certainly $G \cong Z_p$ is cyclic and $\psi(d) = \varphi(d)$ for each $d$ dividing $p$.

Suppose that every subgroup of order $p^t$ is cyclic for some $k \geq t \geq 1$ and that for all such $t$, $\psi(p^t) = \varphi(p^t)$. Let $G$ be a subgroup of order $p^{k+1}$. Note that the elements of $G$ are precisely the roots of $x^{p^{k+1}} - 1$ in $F$. Similarly, the roots of $x^{p^k} - 1$ are precisely those elements of $G$ whose order divides $p^k$. Suppose $\alpha$ and $\beta$ are two such roots; that is, $\alpha^{p^k} = 1$ and $\beta^{p^k} = 1$. Now $(\beta^{-1})^{p^k} = 1$, so that $(\alpha\beta^{-1})^{p^k} = 1$. Since $1^{p^k} = 1$, the set of roots of $x^{p^k} - 1$ (i.e. the set of elements of $G$ whose order divides $p^k$) form a subgroup of $F^\times$ of order $p^k$. By the induction hypothesis, this group is cyclic, and $\psi(p^t) = \varphi(p^t)$ for each $1 \leq t \leq k$. Note also that for each such $t$, $\psi(p^t)$ with respect to $G$ is equal to $\psi(p^t)$ with respect to this subgroup. Since $\sum_{t=1}^{k+1} \psi(p^t) = p^t = \sum_{t=1}^{k+1} \varphi(p^t)$, we have $\psi(p^{k+1}) = \varphi(p^{k+1})$. In particular, $G$ contains an element of order $p^{k+1}$, and so is cyclic.

Thus every finite $p$-subgroup of $F^\times$ is cyclic. Now let $G$ be an arbitrary finite subgroup; since $G$ is abelian, it is the direct product of its $p$-subgroups. Since these are cyclic, $G$ is cyclic.

Subgroups of finitely generated groups need not be finitely generated

Prove that the commutator subgroup of the free group on 2 generators is not finitely generated. (In particular, subgroups of finitely generated groups need not be finitely generated.)

We begin with a lemma.

Lemma: A free group of countably infinite rank is not finitely generated. Proof: Let $A$ be a countably infinite set and suppose $F(A)$ is finitely generated, say by $S$. Now only a finite number of elements of $A$ appear as letters in the words in $S$. Thus some word in $F(A)$ is not in $\langle S \rangle$, a contradiction. $\square$

Let $F_2 = F(a,b)$ be the free group of rank 2 and let $C = \{ [x,y] \ |\ x,y \in F_2 \}$ be the set of commutators in $F_2$. By the universal property of free groups and using the inclusion $C \rightarrow F_2^\prime$, there exists a unique group homomorphism $\Phi : F(C) \rightarrow F_2^\prime$, which is surjective by construction. Now suppose $w \in \mathsf{ker}(\Phi)$. Now $\Phi(w) = 1$ is a word in $F_2^\prime \leq F_2$; since this group is free on $a$ and $b$, we must have $w = 1$. Thus $\Phi$ is injective, and we have $F_2^\prime \cong F(C)$. Hence $F_2^\prime$ is a free group of infinite rank. By the lemma, $F_2^\prime$ is not finitely generated.

The class of nilpotent groups is closed under subgroups and quotients, but not extensions

Prove that subgroups and quotient groups of nilpotent groups are nilpotent. (Your proof should work for infinite groups.) Give an explicit example of a group $G$ which possesses a normal subgroup $H$ such that $H$ and $G/H$ are nilpotent but $G$ is not nilpotent.

We showed that subgroups and quotients of nilpotent groups are nilpotent in the lemmas to this previous exercise.

Consider now $D_6$. $\langle r \rangle \leq D_6$ is a normal subgroup of order 3, and thus $\langle r \rangle \cong Z_3$ is abelian, hence nilpotent. Moreover, $D_6/\langle r \rangle \cong Z_2$ is abelian, hence nilpotent. However, $D_6$ has distinct Sylow 2-subgroups $\langle s \rangle$ and $\langle sr \rangle$ (for example). By Theorem 3, $D_6$ is not nilpotent.

If the quotients of a group by two subgroups are abelian, then the quotient by their intersection is abelian

Let $A,B \leq G$ be normal subgroups such that $G/A$ and $G/B$ are abelian. Prove that $G/(A \cap B)$ is abelian.

Since $G/A$ is abelian we have $[G,G] \leq A$. Likewise, $[G,G] \leq B$. Thus $[G,G] \leq A \cap B$, so that $G/(A \cap B)$ is abelian.

The commutator generated subgroup operator is commutative

Let $G$ be a group. Prove that if $x,y \in G$ then $[y,x] = [x,y]^{-1}$. Deduce that for any subsets $A,B \subseteq G$, $[A,B] = [B,A]$. (Recall that $[A,B]$ is the subgroup of $G$ generated by the commutators $[a,b]$.)

Note that $[x,y][y,x] = (x^{-1}y^{-1}xy)(y^{-1}x^{-1}yx) = 1$. By the uniqueness of inverses, $[x,y]^{-1} = [y,x]$.

Now $[A,B] = \langle S \rangle$, where $S = \{ [a,b] \ |\ a \in A, b \in B \}$. Now $S^{-1} = \{ [a,b]^{-1} \ |\ a \in A, b \in B \}$ $= \{ [b,a] \ |\ a \in A, b \in B \}$, so that $[A,B] = \langle S \rangle$ $= \langle S^{-1} \rangle$ $= [B,A]$.

Exhibit the distinct cyclic subgroups of an elementary abelian group of order p²

Let $p$ be a prime. Let $A$ and $B$ be cyclic groups of order $p$ with generators $x$ and $y$, respectively. Set $E = A \times B$ so that $E$ is the elementary abelian group of order $p^2$. Prove that the distinct subgroups of $E$ of order $p$ are $\langle (x,1) \rangle$, $\langle (x,y) \rangle$, …, $\langle (x,y^{p-1}) \rangle$, $\langle (1,y) \rangle$.

We saw previously (by counting) that $E$ has precisely $p+1$ distinct subgroups of order $p$. Thus it suffices to show that these subgroups are mutually distinct.

Suppose $\langle (x^a,y^b) \rangle = \langle (x^c,y^d) \rangle$. Then $(x^c,y^d) = (x^a,y^b)^k$ for some $k$, hence $(x^{ak},y^{bk}) = (x^c,y^d)$. Thus $x^{ak} = x^c$, so that $ak \equiv c$ mod $p$. Likewise, $bk \equiv d$ mod $p$.

Suppose $a \equiv 1$. Then $c \not\equiv 0$ since otherwise the first coordinate of each element of $\langle (x^c,y^d) \rangle$ is 1, a contradiction since $a \neq 1$. Thus none of the first $p$ subgroups above is equal to the last. If (without loss of generality, considering the generators chosen above) $c \equiv 1$, then $k \equiv 1$, and we have $b \equiv d$ mod $p$. Thus the first $p$ subgroups in the list above are mutually distinct. Hence the subgroups given above exhaust the distinct order $p$-subgroups of $E$.

Exhibit Sym(n) as a subgroup of a general linear group

Let $G_i = F$ be a field for all $i$ and use the preceding exercise to show that the set of $n \times n$ matrices over $F$ such that each row and column contains exactly one 1 and all other entries are 0 is a subgroup of $GL_n(F)$ isomorphic to $S_n$. (These matrices are called permutation matrices since they simply permute the standard basis $e_1,\ldots,e_n$ (as above) of the $n$-dimensional vector space $F^n$.)

We know that if $F$ is a finite field then $\mathsf{Aut}(F^n) \cong GL_n(F)$. This isomorphism $\zeta$ can be defined as follows: given $\theta \in \mathsf{Aut}(F^n)$, $\zeta(\theta)$ is the matrix in $GL_n(F)$ whose $i$th row is precisely $\theta(e_i)$. (In particular, $\zeta$ is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In the previous exercises, we found an injective group homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(F^n)$. Combining these results we have an injective group homomorphism $\Psi : S_n \rightarrow GL_n(F)$, computed as follows: $\Psi(\pi) = [e_{\pi(1)}\ \cdots\ e_{\pi(n)}]^T$. We can see that each $\Psi(\pi)$ is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus $S_n$ is identified with a subgroup of $GL_n(F)$ consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.

Thus we have proven the result for a finite field $F$; we began with this case because it was shown previously that $\mathsf{Aut}(F^n) \cong GL_n(F)$. If this is true for arbitrary fields then the same proof carries over to arbitrary fields $F$; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields $F$ by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not $k = 0$ in $F$ for some integer $k$ never arises, so that all computations hold for arbitrary fields (in which $1 \neq 0$) and in fact the set of permutation matrices over any $F$ is closed under matrix multiplication.

In the direct product of the quaternion group and an elementary abelian 2-group, all subgroups are normal

Prove that all subgroups of $Q_8 \times E_{2^k}$ are normal, where $E_{2^k} = Z_2 \times \cdots \times Z_2$ is the elementary abelian group of order $2^k$.

Let $H \leq Q_8 \times E_{2^k}$ be a subgroup, and let $\alpha = (a,x) \in H$ and $\beta = (b,y) \in Q_8 \times E_{2^k}$. Since $E_{2^k}$ is abelian, we have $\beta\alpha\beta^{-1} = (bab^{-1}, x)$.

We saw previously that the conjugacy classes of $Q_8$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, and $\{k,-k\}$. In particular, we have $bab^{-1} \in \{a, a^3\}$.

If $bab^{-1} = a$, then $\beta\alpha\beta^{-1} = (a,x) \in H$. If $bab^{-1} = a^3$, then (since every element of $E_{2^k}$ has order 2) $\beta\alpha\beta^{-1} = (a^3,x)$ $= (a^3, x^3)$ $= (a,x)^3 \in H$. Thus $H$ is normal.

Elements of generalized coordinate subgroups commute pairwise

Under the notation of the previous example, let $I$ and $K$ be any disjoint nonempty subsets of $\{1,2,\ldots,n\}$ and let $G_I$ and $G_K$ be the subgroups of $G$ defined above. Prove that $xy = yx$ for all $x \in G_I$ and $y \in G_K$.

Let $x = (x_j)$ and $y = (y_j)$.

If $j \in I$, then $(xy)_j = x_j y_j$ $= x_j \cdot 1$ $= 1 \cdot x_j$ $= y_jx_j$ $= (yx)_j$.

If $j \in K$, then $(xy)_j = x_j y_j$ $= 1 \cdot y_j$ $= y_j \cdot 1$ $= y_jx_j$ $= (yx)_j$.

Otherwise, $(xy)_j = x_j y_j$ $= 1 \cdot 1$ $= y_j \cdot x_j$ $= (yx)_j$.

Thus $xy = yx$.

Generalized coordinate subgroups of a direct product

Let $G_1,G_2,\ldots,G_n$ be groups and let $G = \times_{i=1}^k G_i$. Let $I$ be a proper nonempty subset of $\{1,2,\ldots,n\}$ and let $J = \{1,2,\ldots,n\} \setminus I$. Define $G_I$ to be the set of elements $(g_i)_{i=1}^n$ in $G$ such that $g_j = 1$ for all $j \in J$.

1. Prove that $G_I$ is isomorphic to $\times_{i \in I} G_i$.
2. Prove that $G_I$ is a normal subgroup of $G$ and that $G/G_I \cong G_J$.
3. Prove that $G \cong G_I \times G_J$.

1. Define a mapping $\varphi : G_I \rightarrow \times_{i \in I} G_i$ by $(\varphi(g))_i = \pi_i(g)$. We need to show that $\varphi$ is a bijective homomorphism.
1. Homomorphism: Let $g = (g_i), h = (h_i) \in G_I$. Then for each $i \in I$, $(\varphi(gh))_i = (\varphi((g_j)(h_j)))_i$ $= (\varphi((g_jh_j)))_i$ $= g_ih_i$ $= (\varphi((g_j)))_i (\varphi((h_j)))_i$ $= (\varphi(g) \varphi(h))_i$. Hence $\varphi(gh) = \varphi(g) \varphi(h)$, so that $\varphi$ is a homomorphism.
2. Surjective: Let $(g_i)_I \in \times_{i \in I} G_i$. Define $(a_i) \in G_I$ by $a_i = g_i$ if $i \in I$ and $a_i = 1$ otherwise. Clearly this is indeed an element of $G_I$. Moreover, $\varphi((a_j)) = (g_j)$. Because $(g_i)$ is arbitrary, $\varphi$ is surjective.
3. Injective: Let $g = (g_i), h = (h_i) \in G_I$. Suppose $\varphi(g) = \varphi(h)$. By definition, $g_j = h_j = 1$ for all $j \notin I$. Moreover, for $i \in I$, we have $g_i = (\varphi(g))_i = (\varphi(h))_i = h_i$. Hence $g = h$, so that $\varphi$ is injective.
2. Let $(g_i) \in G_I$ and $(x_i) \in G$. Note that $(x_i)^{-1} (g_i) (x_i) = (x_i^{-1} g_i x_i)$. For $i \in I$, we have $g_i = 1$, so that $x_i^{-1} g_i x_i = 1$. Thus $(x_i)^{-1} (g_i) (x_i) \in G_I$. Hence $G_I \leq G$ is normal.

Now define a mapping $\psi : G \rightarrow G_J$ as follows: $(\psi((g_i)))_j = g_j$ if $j \in J$ and $1$ otherwise. We wish to show that $\psi$ is a surjective homomorphism and that $\mathsf{ker}\ \psi = G_I$.

1. Homomorphism: Let $g = (g_i), h = (h_i) \in G$. If $i \in J$, then $\psi(gh)_i = \psi((g_jh_j))_i = g_ih_i$ $= \psi(g)_i \psi(h)_i$. If $i \notin J$, then $\psi(gh)_i = 1 = 1 \cdot 1$ $= \psi(g)_i \psi(h)_i$. Thus $\psi$ is a homomorphism.
2. Surjective: It is clear that if $g \in G_J$, then $\psi(g) = g$.
3. Kernel: $(\subseteq)$ Let $g = (g_i) \in \mathsf{ker}\ \psi$. Then $(\psi(g_j))_i = 1$ for all $i$; in particular, $g_i = 1$ if $i \notin I$. Thus $g \in G_I$. $(\supseteq)$ Let $g = (g_i) \in G_I$. Now if $j \in J$, $g_j = 1$. Clearly then $\psi(g) = 1$, so that $g \in \mathsf{ker}\ \psi$.

By the First Isomorphism Theorem, we have $G/G_I \cong G_J$.

3. Define a mapping $\theta : G \rightarrow G_I \times G_J$ as follows. $(\theta(g)_1)_i = g_i$ for $i \in I$ and $1$ otherwise, and $(\theta(g)_2)_j = g_j$ for $j \in J$ and $1$ otherwise. We wish to show that $\theta$ is a bijective homomorphism.
1. Homomorphism: Let $g = (g_k), h = (h_k) \in G$. If $i \in I$, then $(\theta((g_k)(h_k))_1)_i = (\theta((g_kh_k))_1)_i$ $= g_ik_i$ $= (\theta((g_k))_1)_i (\theta((h_k))_1)_i$. If $i \notin I$, then $(\theta((g_k)(h_k))_1)_i = (\theta((g_kh_k))_1)_i$ $= 1 = 1 \cdot 1$ $= (\theta((g_k))_1)_i (\theta((h_k))_1)_i$. Thus $\theta(gh)_1 = \theta(g)_1 \theta(h)_1$. Similarly, $\theta(gh)_2 = \theta(g)_2 \theta(h)_2$. Hence $\theta(gh) = \theta(g) \theta(h)$, so that $\theta$ is a homomorphism.
2. Surjective: Let $(a,b) \in G_I \times G_J$, with $a = (a_k)$ and $b = (b_k)$. Define $g = (g_k) \in G$ as follows: If $k \in I$ then $g_k = a_k$, and if $k \in J$ then $g_k = b_k$. Then by definition, we have $(\theta(g)_1)_k = a_k$ if $k \in I$ and $1$ otherwise, so that $\theta(g)_1 = a$. Similarly, $\theta(g)_2 = b$. Hence $\theta(g) = (a,b)$, and $\theta$ is surjective.
3. Injective: Suppose $g = (g_k), h = (h_k) \in G$ such that $\theta(g) = \theta(h)$. For $k \in I$, we have $g_k = (\theta(g)_1)_k$ $= (\theta(h)_1)_k = h_k$, and for $k \in J$, we have $g_k = (\theta(g)_2)_k$ $= (\theta(h)_2)_k = h_k$. Thus $g = h$, and $\theta$ is injective.