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 ith 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.