Tag Archives: cyclic group

Exhibit some finite fields

Let g(x) = x^2+x-1 and h(x) = x^3-x+1. Construct finite fields of order 4, 8, 9, and 27 elements. For the fields with 4 and 9 elements, give the multiplication tables and show that the nonzero elements form a cyclic group.


Note that to construct fields of the given sizes, it suffices to show that g(x) and h(x) are each irreducible over the fields \mathbb{F}_2 and \mathbb{F}_3. (More precisely, that these polynomials are irreducible when we inject the coefficients into the respective fields.)

It is easy to see that neither g nor h has a root in either \mathbb{F}_2 or in \mathbb{F}_3, so these are indeed irreducible. Then \mathbb{F}_2[x]/(g(x)), \mathbb{F}_2[x]/(h(x)), \mathbb{F}_3[x]/(g(x)), and \mathbb{F}_3[x]/(h(x)) are fields of order 4, 8, 9, and 27 by Proposition 11 in D&F, and specifically are isomorphic as fields to the extension of \mathbb{F}_2 or \mathbb{F}_3 by adjoining a root of g or h.

The multiplication table of \mathbb{F}_2(\alpha) \cong \mathbb{F}_2[x]/(g(x)) is as follows.

\begin{array}{c|cccc} & 0 & 1 & \alpha & \alpha+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & \alpha+1 \\ \alpha & 0 & \alpha & \alpha+1 & 1 \\ \alpha+1 & 0 & \alpha+1 & 1 & \alpha \end{array}

Evidently, (\mathbb{F}_2(\alpha))^\times is generated by \alpha.

The multiplication table of \mathbb{F}_3(\alpha) \cong \mathbb{F}_3[x]/(g(x)) is as follows.

\begin{array}{c|ccccccccc} & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ 2 & 0 & 2 & 1 & 2\alpha & 2\alpha+2 & 2\alpha+1 & \alpha & \alpha+2 & \alpha+1 \\ \alpha & 0 & \alpha & 2\alpha & 2\alpha+1 & 1 & \alpha+1 & \alpha+2 & 2\alpha+2 & 2 \\ \alpha+1 & 0 & \alpha+1 & 2\alpha+2 & 1 & \alpha+2 & 2\alpha & 2 & \alpha & 2\alpha+1 \\ \alpha+2 & 0 & \alpha+2 & 2\alpha+1 & \alpha+1 & 2\alpha & 2 & 2\alpha+2 & 1 & \alpha \\ 2\alpha & 0 & 2\alpha & \alpha & \alpha+2 & 2 & 2\alpha+2 & 2\alpha+1 & \alpha+1 & 1 \\ 2\alpha+1 & 0 & 2\alpha+1 & \alpha+2 & 2\alpha+2 & \alpha & 1 & \alpha+1 & 2 & 2\alpha \\ 2\alpha+2 & 0 & 2\alpha+2 & \alpha+1 & 2 & 2\alpha+1 & \alpha & 1 & 2\alpha & \alpha+2 \end{array}

Evidently, (\mathbb{F}_3(\alpha))^\times is generated by \alpha.

Compute Hom(ZZ/(m), ZZ/(n)) over ZZ

Compute \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n)).


We claim that \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong_\mathbb{Z} \mathbb{Z}/(m,n). (We use \cong_R to mean “isomorphic as R-modules”.)

By this previous exercise, we have \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong \mathsf{Ann}_{\mathbb{Z}/(n)}((m)).

Let d be a greatest common divisor of m and n, and write m = ad and n = bd where a and b are relatively prime. Note that m \cdot \overline{b} = a \cdot (d \cdot \overline{b}) = a \cdot (\overline{bd}) = a \cdot \overline{0} = \overline{0}. So \overline{b} \in \mathsf{Ann}_{\mathbb{Z}/(n)}((m)). Now define \varphi : \mathbb{Z} \rightarrow \mathsf{Ann}_{\mathbb{Z}/(n)}((m)) by k \mapsto k \cdot \overline{b}; certainly \varphi is a \mathbb{Z}-module homomorphism. We claim that \varphi is surjective. To see this, suppose \overline{t} \in \mathsf{Ann}_{\mathbb{Z}/(n)}((m)). In particular, m \cdot \overline{t} = \overline{mt} = \overline{0}, so that n|tm. Then b|at, and since a and b are relatively prime, by Euclid’s lemma, we have b|t. So t = bs for some s. Thus \overline{t} = \varphi(s), and \varphi is surjective. Next we claim that \mathsf{ker}\ \varphi = (m,n) = (d). Certainly \varphi(d) = d \cdot \overline{b} = \overline{0}, so that (d) \subseteq \mathsf{ker}\ \varphi. Now suppose s \in \mathsf{ker}\ \varphi, so that \varphi(s) = s \cdot \overline{b} = \overline{sb} = \overline{0}. Then n|sb, so that d|s, and we have s \in (d) as desired. By the First Isomorphism Theorem, we have \mathsf{Ann}_{\mathbb{Z}/(n)}((m)) \cong \mathbb{Z}/(m,n).

Thus we have \mathsf{Hom}_{\mathbb{Z}}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong_R \mathbb{Z}/(m,n).

Find a generating set for the augmentation ideal of a group ring

Let R be a commutative ring with 1 \neq 0 and G a finite group. Prove that the augmentation ideal in the group ring R[G] is generated by \{ g-1 \ |\ g \in G \}. Prove that if G = \langle \sigma \rangle is cyclic then the augmentation ideal is generated by \sigma - 1.


Recall that the augmentation ideal of R[G] is the kernel of the ring homomorphism R[G] \rightarrow R given by \sum r_ig_i \mapsto \sum r_i; that is, it consists of all elements in R[G] whose coefficients sum to 0 in R.

Let S = \{ g-1 \ |\ g \in G \}, and let A denote the augmentation ideal of R[G]. First, note that g-1 \in A for each g \in G, so that (g-1) \subseteq A. Thus (S) \subseteq A. Now let \alpha = \sum_{g_i \in G} r_ig_i \in A; then we have \sum r_i = 0. Consider the following.

\sum_{g_i \in G} r_i(g_i - 1)  =  \sum_{g_i \in G} r_ig_i - r_i
 =  = \left( \sum_{g_i \in G} r_ig_i \right) - (\sum_{g_i \in G} r_i)
 =  \alpha - 0
 =  \alpha

Thus A \subseteq (S), and we have A = (S).

Suppose further that G = \langle \sigma \rangle is cyclic. We still have (\sigma - 1) \subseteq A. Now note that (\sigma - 1)\left( \sum_{t=0}^k \sigma^t \right) = \sigma^{k+1} - 1, so that \sigma^k - 1 \in (\sigma - 1) for all k. Thus A = (\sigma - 1).

Characterization of free abelian groups of finite rank

Let S be a set. The group presented by (S, R) where R = \{ [s,t] \ |\ s,t \in S \} is called the free abelian group on S. Denote it by A(S). Prove that A(S) has the following universal property: If G is any abelian group and \varphi S \rightarrow G is a set function, then there is a unique group homomorphism \Phi : A(S) \rightarrow G such that \Phi|_S = \varphi. Deduce that if A is a free abelian group on a set of (finite) cardinality n, then A \cong \mathbb{Z}^n.


Recall from the universal property of free groups that there is a unique group homomorphism \Psi : F(S) \rightarrow G such that \Psi|_S = \varphi. Moreover, clearly R \leq \mathsf{ker}\ \Psi. Then there is a unique group homomorphism \Psi : A(S) \rightarrow G such that \Psi = \Phi \circ \pi. Identifying s in S with sR in A(S), we have \Phi|_S = \varphi. This gives existence.

To see uniqueness, note that any homomorphism \Phi : A(S) \rightarrow G extending \varphi is uniquely determined by \Psi, which is uniquely determined by \varphi.

Now let n \in \mathbb{N} and let S = \{1, \ldots, n \}. If n = 0, then A(S) is a quotient of F(S) = 1, so that in fact A(S) = 1. Suppose n > 0, and for each 1 \leq k \leq n, define \varphi(k) to be the n-tuple with 1 in the kth coordinate and 0 in all other coordinates. By the universal property, there exists a unique group homomorphism \Phi : A(S) \rightarrow \mathbb{Z}^n extending \varphi; \Phi is surjective since \Phi[S] generates \mathbb{Z}^n. Now if w = s_1^{e_1} \cdots s_n^{e_n} \in \mathsf{ker}\ \Phi, note that since A(S) is abelian, we can collect terms so that s_i = i. Now \Phi(w) = (e_1, \ldots, e_n) = (0, \ldots, 0), and in fact w = 1. Thus \Phi is injective, and we have A(S) \cong \mathbb{Z}^n.

The center of a subgroup of index 2 in a 2-group with cyclic center has rank at most 2

Use the preceding exercise to prove that if P is a 2-group which has a cyclic center and M is a subgroup of index 2 in P, then Z(M) has rank at most 2.


Note that M \leq P is normal and that Z(M) \leq M is characteristic; thus Z(M) is normal in P. Moreover, letting Z(M)_2 = \{ z \in Z(M) \ |\ z^2 = 1 \}, we know that Z(M)_2 \leq Z(M) is characteristic, so that Z(M)_2 \leq P is normal. Moreover, Z(M)_2 is elementary abelian by construction. Note that, by §5.2 #7, Z(M) and Z(M)_2 have the same rank.

If w \in P, then conjugation by w is an automorphism of Z(M)_2. If we choose w \notin M, and let z be fixed by w– that is, wz = zw, then we have z \in C_P(M) (since z \in Z(M)) and z \in C_P(\langle w \rangle). Since P = M \langle w \rangle, in fact w \in Z(P). Thus the subgroup of Z(M)_2 which is fixed under conjugation by w is contained in Z(P), and thus has rank 1.

By this previous exercise, Z(M)_2 has rank at most 2, so that Z(M) also has rank at most 2.

If the first two lower central quotients of a group are cyclic, then the second derived subgroup is trivial

Show that if G^\prime/G^{\prime\prime} and G^{\prime\prime} / G^{\prime\prime\prime} are both cyclic then G^{\prime\prime} = 1.


We begin by proving a lemma.

Lemma: Let G be a group and let N \leq G be normal. Then (G/N)^\prime = G^\prime N /N. Proof: (\subseteq) If [xN,yN] \in G/N is a commutator, then [xN,yN] = [x,y]N \in G^\prime N/N. (\supseteq) G^\prime \leq G^\prime N, so that G/G^\prime N \cong (G/N)/(G^\prime N/N) is abelian. By Theorem 5.7 in the text, (G/N)^\prime \leq G^\prime N/N. \square

Now for the main result.

Note that G^{\prime\prime\prime} \leq G is characteristic, hence normal, so that G^{\prime\prime}/G^{\prime\prime\prime} \leq G/G^{\prime\prime\prime} is a cyclic, hence abelian, normal subgroup.

By this previous exercise, (G/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}) \cong G/G^{\prime\prime} (via the Third Isomorphism Theorem) acts on G^{\prime\prime}/G^{\prime\prime\prime} by conjugation on the left as follows: (gG^{\prime\prime}) \cdot (a G^{\prime\prime\prime}) = (gag^{-1}) G^{\prime\prime\prime}. Now conjugation is an automorphism of G^{\prime\prime}/G^{\prime\prime\prime}, and since this quotient is cyclic, its automorphism group is abelian. Thus, for all g,h \in G and a \in G^{\prime\prime}, we have (gh)a(gh)^{-1} G^{\prime\prime\prime} = (hg)a(hg)^{-1} G^{\prime\prime\prime}. Via some arithmetic, we see that [g,h] a [g,h]^{-1} a^{-1} \in G^{\prime\prime\prime}; in particular, each element of G^{\prime\prime}/G^{\prime\prime\prime} commutes with each generator of G^{\prime}/G^{\prime\prime\prime}, so that G^{\prime\prime}/G^{\prime\prime\prime} \leq Z(G^{\prime}/G^{\prime\prime\prime}).

Using the Third Isomorphism Theorem, (G^\prime/G^{\prime\prime\prime})/Z(G^\prime/G^{\prime\prime\prime}) \cong ((G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime})) \cong (G^\prime/G^{\prime\prime})/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime})) is cyclic; hence G^\prime/G^{\prime\prime\prime} is abelian, and using the lemma we have 1 = (G^\prime/G^{\prime\prime\prime})^\prime = G^{\prime\prime}/G^{\prime\prime\prime}. Thus G^{\prime\prime} = G^{\prime\prime\prime}.

An equivalent characterization of finite cyclic groups

Prove the following proposition by invoking the Fundamental Theorem of Finite Abelian Groups.

Proposition: Let G be a finite abelian group such that, for all positive integers n dividing |G|, G contains at most n elements satisfying x^n = 1. Prove that G is cyclic.


Proof: By FTFGAG, we have G \cong Z_{n_1} \times \cdots \times Z_{n_t} for some integers n_i such that n_{i+1}|n_i for all i. Suppose t \geq 2, and let p be a prime divisor of n_2. Now p also divides n_1. Now Z_{n_2} contains an element x of order p, and also Z_{n_1} contains an element y of order p. Now every element z \in \langle y \rangle \times \langle x \rangle satisfies z^p = 1, and there are p^2 such elements, a contradiction. Thus t = 1, hence G = Z_{n_1} is cyclic.

Some properties of nonabelian p groups of order p³

Prove that if p is a prime and P a nonabelian group of order p^3, then |Z(P)| = p and P/Z(P) \cong Z_p \times Z_p.


By Lagrange, there are 4 possibilities for |Z(P)|: 1, p, p^2, and p^3. However, we know that Z(P) is nontrivial, and if |Z(P)| = p^3, then P is abelian.

If |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, and we have P abelian, a contradiction.

If |Z(P)| = p, there are two possibilities for the isomorphism type of P/Z(P): Z_{p^2} and Z_p \times Z_p. In the first case, P is abelian. Thus P/Z(P) \cong Z_p \times Z_p.

Equivalent characterizations of nilpotent and cyclic groups

If G is finite, prove that (1) G is nilpotent if and only if it has a normal subgroup of each order dividing |G|, and (2) is cyclic if and only if it has a unique subgroup of each order dividing |G|.


We begin with some lemmas.

Lemma 1: Let G be a group, and let H_1, \ldots, H_k be characteristic subgroups of G. Then H = H_1\cdots H_k is characteristic in G. Proof: H is a subgroup because each H_i is normal. If \alpha is an automorphism of G, then \alpha[H] = \alpha[H_1] \cdots \alpha[H_k] = H_1 \cdots H_k = H. \square

Lemma 2: Let G be a finite group. If P \leq G is a normal Sylow subgroup, then P is characteristic. Proof: P is the unique subgroup of order |P|. \square

Now to the main result; first we prove (1).

(\Rightarrow) We proceed by induction on the breadth (number of prime divisors with multiplicity) of G.

For the base case, if G is a finite nilpotent group of breadth 1, then G \cong Z_p is an abelian simple group. Thus G has a normal subgroup of order 1 and p, the only divisors of |G|.

For the inductive step, suppose that every finite nilpotent group of breadth k \geq 1 has a normal subgroup of each order dividing |G|, and let G be a finite nilpotent group of breadth k+1. Now Z(G) is nontrivial, so there exists an element x \in Z(G) of prime order, and \langle x \rangle G is normal. We showed in a lemma to the previous exercise that G/\langle x \rangle is nilpotent, and has breadth k. By the inductive hypothesis, G/\langle x \rangle has a normal subgroup of each order d dividing |G/\langle x \rangle|. By the Lattice Isomorphism Theorem, a subgroup \overline{H} \leq G/\langle x \rangle is normal if and only if \overline{H} = H/\langle x \rangle for some normal subgroup H \leq G. Thus we see that G has a normal subgroup of order d for all d dividing |G| such that p|d. Now let d divide |G|, with p not dividing d, and let H \leq G be a normal subgroup of order pd. Since H \leq G, H is a finite nilpotent group. By Theorem 3, H is the direct product of its Sylow subgroups. Write H = P \times Q, where P is the Sylow p subgroup of H and Q the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of Q is characteristic in H, so that Q is characteristic in H. Note that |Q| = d, thus Q \leq G is normal and has order d.

(\Leftarrow) Suppose G has a normal subgroup of each order d dividing |G|. In particular, G has a normal (thus unique) Sylow p-subgroup for each prime p. By Theorem 6.3, G is nilpotent.

Now we prove (2).

The (\Rightarrow) direction was proved in Theorem 2.7.

(\Leftarrow) Suppose G is finite and has a unique subgroup of order d for each d properly dividing |G|. In particular, all the Sylow subgroups of G are unique, hence normal; say there are k of these. Suppose some Sylow p-subgroup P of G is not cyclic; then by this previous exercise, P has a subgroup congruent to Z_p \times Z_p, and thus G has distinct subgroups of order p, a contradiction. Thus each Sylow subgroup of G is cyclic; say P_i = \langle x_i \rangle. Then |x_1 \cdots x_k| = |G|, so that G = \langle x_1 \cdots x_k \rangle is cyclic.

Construct all the semidirect products of Cyc(2ⁿ) by Cyc(2)

Show that for any n \geq 3 there are exactly 4 distinct homomorphisms from Z_2 into \mathsf{Aut}(Z_{2^n}). Prove that the resulting semidirect products are nonisomorphic groups of order 2^{n+1}. (These four groups, with the cyclic group and the generalized quaternion group, are all the groups of order 2^{n+1} which have a cyclic subgroup of index 2.)


We begin with two lemmas.

Lemma 1: Let k \geq 3 be an integer. Suppose m and n are integers such that, for all integers a and b, m+a(1+2^{k-1})^n \equiv a+m(1+2^{k-1})^b mod 2^k. Then m \equiv n \equiv 0 mod 2. Proof: Note that (1+2^{k-1})^2 \equiv 1 mod 2^k. Suppose n \equiv 1 mod 2. Let b = 2. Then we have a2^{k-1} \equiv 0 mod 2^k, a contradiction when a is odd. Thus n \equiv 0 mod 2^k. Now choosing b odd, we have m2^{k-1} \equiv 0 mod 2^k; thus m \equiv 0 mod 2. \square

Lemma 2: Let k \geq 3 be an integer. Suppose m and n are integers such that, for all integers a and b, m+a(-1+2^{k-1})^n \equiv a+m(-1+2^{k-1})^b mod 2^k. Then m \equiv 0 mod 2^{k-1} and n \equiv 0 mod 2. Proof: Note that (-1+2^{k-1})^2 \equiv 1 mod 2^k. Suppose n \equiv 1 mod 2. Let b = 2. Then we have 0 \equiv 2a(1-2^{k-2}) mod 2^k. Since 1-2^{k-2} is odd, we have 2a \equiv 0 mod 2^k for all a, which is absurd. Thus n \equiv 0 mod 2. Now let b be odd; we have 2m(1-2^{k-2}) \equiv 0 mod 2^k; again since 1-2^{k-2} is odd we have 2m \equiv 0 mod 2^k, so that m \equiv 0 mod 2^{k-1}. \square

We found that there are 4 distinct homomorphisms Z_2 \rightarrow \mathsf{Aut}(Z_{2^n}) in two lemmas to the previous exercise. Moreover, writing Z_2 = \langle x \rangle and Z_{2^n} = \langle y \rangle, these are given as follows. \varphi_1(x)(y) = y, \varphi_2(x)(y) = y^{-1}, \varphi_3(x)(y) = y^{1+2^{n-1}}, and \varphi_4(x)(y) = y^{-1+2^{n-1}}.

Let G_i = Z_{2^n} \rtimes_{\varphi_i} Z_2. We will now show that these G_i are distinct.

Note that \varphi_1 is trivial, so that G_1 \cong Z_{2^n} \times Z_2 is abelian. However, the remaining groups are not abelian, so G_1 is distinct from the remaining three.

Note that G_2 \cong D_{2^{n+1}} = \langle r,s \rangle. By this previous exercise, Z(D_{2^{n+1}}) = \langle r^{2^{n-1}} \rangle \cong Z_2.

Now G_3 is generated by x and y, and we have xyx^{-1} = \varphi_3(x)(y) = y^{1+2^{n-1}}. Thus this group has the presentation \langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{1+2^{n-1}}x \rangle. Similarly, G_4 has the presentation \langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{-1+2^{n-1}}x \rangle.

We now compute the centers of G_3 and G_4.

Every element of G_3 can be written as y^ix^j where 0 < i \leq 2^n and 0 < j \leq 2, and there are 2^{n+1} such expressions. Since we know by other means that |G_3| = 2^{n+1}, every element of G_3 can be written in this form uniquely. Let s = y^ax^b \in Z(G_3) and let y^ix^j \in G_3. Then y^ax^by^ix^j = y^{a+i(1+2^{n-1})^b}x^{b+j} is equal to y^ix^jy^ax^b = y^{i+a(1+2^{n-1})^j}x^{b+j}. Comparing exponents and using Lemma 1, we have a \equiv b \equiv 0 mod 2. Thus s \in \langle y^2 \rangle. Now note that y^2y^ix^j = y^{i+2}x^j and y^ixy^2 = y^iy^{2(1+2^{n-1})}x = y^{i+2}x, so that y^2 \in Z(G_3); thus Z(G_3) = \langle y^2 \rangle \cong Z_{2^{n-1}}.

Performing the same analysis on G_4 reveals that Z(G_4) = \langle y^{2^{n-1}} \rangle \cong Z_2.

Thus G_3 is distinct from G_2 and G_4.

We know that G_2 \cong D_{2^{n+1}} has 2^n + 1 elements of order 2. Suppose s = y^ax^b \in G_4 has order 2; then (y^ax^b)^2 = y^{a(1+(-1+2^{n-1})^b)} = 1. If b = 0, then a = 0, so that s = 1. If b = 1, then a2^{n-1} \equiv 0 mod 2^n, so that a \equiv 0 mod 2. There are 2^{n-1} such numbers between 0 and 2^n. Thus G_4 has 2^{n-1} elements of order 2, and G_2 \not\cong G_4.