Tag Archives: isomorphism

Prove that a given map is a field automorphism

Prove directly that the map \varphi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}) given by a+b\sqrt{2} \mapsto a-b\sqrt{2} is a field isomorphism.


Note that elements of \mathbb{Q}(\sqrt{2}) can be written uniquely in the form a+b\sqrt{2} where a,b \in \mathbb{Q}, so that this rule indeed defines a function.

Now \varphi((a+b\sqrt{2}) + (c+d\sqrt{2})) = \varphi((a+c) + (b+d)\sqrt{2}) = (a+c) - (b+d)\sqrt{2} = (a-b\sqrt{2}) + (c-d\sqrt{2}) = \varphi(a+b\sqrt{2}) + \varphi(c+d\sqrt{2}), so that \varphi preserves addition.

Similarly, we have \varphi((a+b\sqrt{2})(c+d\sqrt{2})) = \varphi((ac+2bd) + (ad+bc)\sqrt{2}) = (ac+2bd) - (ad+bc)\sqrt{2} = (a-b\sqrt{2})(c-d\sqrt{2}) = \varphi(a+b\sqrt{2})\varphi(c+d\sqrt{2}), so that \varphi preserves multiplication.

Thus \varphi is a ring homomorphism, and hence a field homomorphism.

Note that if \varphi(a+b\sqrt{2}) = 0, then we have a-b\sqrt{2} = 0, and so if b = 0 we have a/b = \sqrt{2}, where a and b are rational numbers- a contradiction since \sqrt{2} is not rational. So b = 0, and thus a = 0, and we have \mathsf{ker}\ \varphi = 0. So \varphi is injective.

Finally, \varphi is surjective since we have \varphi(a+(-b)\sqrt{2}) = a+b\sqrt{2} for all a and b.

So \varphi is a field isomorphism.

The tensor product of the quotients of a commutative ring by two ideals is isomorphic to the quotient by their sum

Let R be a commutative ring with ideals I and J. Let R/I and R/J be R-modules (in fact (R,R)-bimodules) in the usual way.

  1. Prove that every element of R/I \otimes_R R/J can be written as a simple tensor of the form (1 + I) \otimes (r + J).
  2. Prove that R/I \otimes_R R/J \cong_R R/(I+J).

We will prove the first result first for simple tensors; the extension to arbitrary sums of tensors follows by tensor distributivity. Let (a+I) \otimes (b+J) be an arbitrary simple tensor in R/I \otimes_R R/J. Now (a+I) \otimes (b+J) = (1+I)a \otimes (b+J) = (1+I) \otimes a(b+J) = (1+I) \otimes (ab+J), as desired.

Now define \varphi : R/I \times R/J \rightarrow R/(I+J) by (a+I, b+J) \mapsto (a+b) + (I+J). First, suppose a_1-a_2 \in I and b_1-b_2 \in J. Then b_1(a_1-a_2) \in I and a_2(b_1-b_2) \in J, so that a_1b_1 - a_2b_2 \in I+J. Thus \varphi is well-defined. It is clear that \varphi is R-balanced, and so induces an R-module homomorphism \Phi : R/I \otimes_R R/J \rightarrow R/(I+J). Since \varphi((1+I) \otimes (r+J)) = r + (I+J), \Phi is surjective. Now suppose (1+I) \otimes (r+J) is in the kernel of \Phi; then r \in I+J. Say r = a+b where a \in I and b \in J. Then (1 + I) \otimes (r + J) = (1+I) \otimes (a+J) + (1+I) \otimes (b+J) = (a+I) \otimes (1+J) + (1+I) \otimes (b+J) = 0+0 = 0, and hence \Phi is an isomorphism.

Binary tensor products essentially commute with arbitrary direct sums

Let R be a ring, let I be a nonempty set, let M be a right R-module, and let \{N_i\}_I be a family of left R-modules. Prove that as abelian groups, M \otimes_R (\bigoplus_I N_i) \cong \bigoplus_I (M \otimes_R N_i).


We will make use of the universal properties of tensor products and direct sums to prove this without too many fine details.

Recall that for each i, we have a canonical injection \iota_i : N_i \rightarrow \bigoplus_I N_i. Thus for all i we have a group homomorphism 1 \otimes \iota_i : M \otimes_R N_i \rightarrow M \otimes_R (\bigoplus_I N_i). By the universal property of direct sums, we have a (unique) group homomorphism \Phi = \bigoplus_I (1 \otimes \iota_i) : \bigoplus_I (M \otimes_R N_i) \rightarrow M \otimes_R (\bigoplus_I N_i) such that \Phi((m \otimes n_i)) = \sum m \otimes \iota_i(n_i).

Now define \psi : M \times \bigoplus_I N_i \rightarrow \bigoplus_I (M \otimes_R N_i) by (m,(n_i)) \mapsto \sum_{n_i \neq 0} m \otimes n_i. This mapping is well defined because for a fixed (n_i), finitely many n_i are nonzero. Evidenly \psi is bilinear, and so induces a group homomorphism \Psi : M \otimes_R (\bigoplus_I N_i) \rightarrow \bigoplus_I (M \otimes_R N_i).

We claim that \Phi and \Psi are mutual inverses. To see this, note that (\Phi \circ \Psi)(m \otimes (n_i)) = \Phi(\sum m \otimes n_i) = m \otimes (n_i) and (\Psi \circ \Phi)((m \otimes n_i)) = \Psi(\sum m \otimes \iota_i(n_i)) = \Psi(m \otimes \sum \iota_i(n_i)) = (m \otimes n_i).

Thus both \Phi and \Psi are group isomorphisms, and we have M \otimes_R (\bigoplus_I N_i) \cong \bigoplus_I (M \otimes_R N_i).

“Is isomorphic to” is an equivalence relation on any set of R-modules

Let R be a ring with 1 and let \mathcal{M} be any set of (left) R-modules. Prove that the relation “is isomorphic to” is an equivalence relation on \mathcal{M}.


Recall that we say M \cong N if there exists an R-module isomorphism \theta : M \rightarrow N. We need to show that this relation is reflexive, symmetric, and transitive.

  1. (Reflexive) Let M \in \mathcal{M}. Note that the identity mapping \mathsf{id} : M \rightarrow M is an R-module isomorphism since \mathsf{id}(x + r \cdot y) = x + r \cdot y = \mathsf{id}(x) + r \cdot \mathsf{id}(y) for all x, y \in M and r \in R. Thus M \cong M, and so \cong is reflexive.
  2. (Symmetric) Suppose M,N \in \mathcal{M} such that M \cong N. That is, there exists an R-module isomorphism \theta : M \rightarrow N. Because \theta is a bijection, the converse relation \theta^{-1} is well-defined, and thus a bijective function N \rightarrow M. We claim that \theta^{-1} is in fact an R-module homomorphism. To see this, let r \in R and x,y \in N. Note that x = \theta(a) and y = \theta(b) for some a,b \in M. Now \theta^{-1}(x + r \cdot y) = \theta^{-1}(\theta(a) + r \cdot \theta(b)) = \theta^{-1}(\theta(a + r \cdot b)) = a + r \cdot b = \theta^{-1}(x) + r \cdot \theta^{-1}(y). So \theta^{-1} : N \rightarrow M is an R-module isomorphism, and we have N \cong M.
  3. (Transitive) Suppose M,N,T \in \mathcal{M} such that M \cong N and N \cong T. Then there exist R-module isomorphisms \theta : M \rightarrow N and \zeta : N \rightarrow T. Recall that \zeta \circ \theta : M \rightarrow T is a bijection; we claim that it is also an R-module homomorphism. To see this, let x,y \in M and r \in R. Then (\zeta \circ \theta)(x + r \cdot y) = \zeta(\theta(x + r \cdot y)) = \zeta(\theta(x) + r \cdot \theta(y)) = \zeta(\theta(x)) + r \cdot \zeta(\theta(y)) = (\zeta \circ \theta)(x) + r \cdot (\zeta \circ \theta)(y). So \zeta \circ \theta : M \rightarrow T is an R-module isomorphism, and we have M \cong T.

Thus \cong is an equivalence relation on \mathcal{M}.

Note also that \mathsf{id} is a unital R-module homomorphism, that if \theta is a unital R-module homomorphism then so is \theta^{-1}, and if \theta and \zeta are unital R-module homomorphisms (and the composition makes sense) then so is \zeta \circ \theta. Thus, if we replace “R-modules” by “unital R-modules”, the result still holds.

Renaming the variables in a polynomial ring does not change its isomorphism type

Let R be a commutative ring and x_1, x_2, … x_n be symbols. Prove that R[x_1,\ldots,x_n] \cong R[x_{\sigma(1)}, \ldots, x_{\sigma(n)}], where \sigma is any permutation in S_n.


We will approach this problem from a slightly different perspective; we will see that polynomial rings are really monoid rings. Given a ring R and a monoid M, the monoid ring R[M] is defined just like the group ring R[G]. (Recall that a monoid is a set with (1) an associative binary operator and (2) an identity element.)

Lemma 1: Let R be a commutative ring. Then the polynomial ring R[x] is isomorphic to the monoid ring R[\mathbb{N}], where \mathbb{N} is considered a monoid under addition. Proof: Define \varphi : R[x] \rightarrow R[\mathbb{N}] as follows: \varphi(\sum_{k=0}^n r_kx^k) = \sum_{k=0}^n r_k \cdot k. It is easy to see that this mapping is an isomorphism. \square

Lemma 2: Let R be a commutative ring and let M,N be monoids. Then R[M \times N] \cong (R[M])[N]. Proof: Define \varphi : R[M \times N] \rightarrow (R[M])[N] by \varphi(\sum_{i,j} r_{i,j}(m_i,n_i)) = \sum_{i,j} (r_{i,j}m_i)n_j, where i and j index M and N, respectively. It is clear that this mapping is surjective. It is also clear that its kernel is trivial, so that \varphi is injective. That \varphi is a ring homomorphism is also clear. \square

Now it is clear that renaming the variables in a polynomial ring (with finitely many variables) does not change its isomorphism type, as in both cases the ring is isomorphic to R[\mathbb{N}^k] for the same k. We could extend this to infinitely many variables by defining the direct sum of monoids in analogous fashion to direct sums of groups- that is, the subset of a direct products in which each element has only finitely many nonidentity entries.

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.

A group of order 12 with no subgroup of order 6 is isomorphic to Alt(4)

Show that a group of order 12 with no subgroup of order 6 is isomorphic to A_4.


Note that 12 = 2^2 \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. Let x \in P_2 have order 2 and y \in P_3 have order 3 by Cauchy; then xy has order 6, a contradiction. Now suppose 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 |N_G(P_2 \cap Q_2)| is divisible by 2^2 and another prime; thus we have P_2 \cap Q_2 normal in G. Recall that every normal group of order 2 in a finite group is central; thus we have an element x of order 2 in the center of G. Let y \in G have order 3 by Cauchy; then xy has order 6, a contradiction.

Thus n_3(G) = 4. Let P_3 \leq G be a Sylow 3-subgroup and let N = N_G(P_3). Now N has index 4 and is self-normalizing by a previous result; in particular, N is not normal in G. Now G acts on G/N by left multiplication, yielding a permutation representation G \rightarrow S_{G/N} whose kernel K is contained in N. Note that |N| = 3, so that either K = 1 or K = N. If K = N, then for all g \in G and h \in N, we have hgN = gN, so that g^{-1}hg \in N. Thus, for all g \in G, gNg^{-1} \leq N; since G is finite, this implies that N \leq G is normal, a contradiction. Thus K = 1 and in fact G \leq S_4. Since G contains 4 \cdot 2 = 8 elements of order 3, it contains all elements of order 3 in S_4. The subgroup generated by the three cycles in S_4 is A_4, so that A_4 \leq G; since these subgroups are finite and have the same cardinality, in fact G \cong A_4.

A criterion for detecting isomorphisms among semidirect products by a cyclic group

Let K be a cyclic group, H an arbitrary group, and \varphi_1 and \varphi_2 homomorphisms K \rightarrow \mathsf{Aut}(H) such that \mathsf{im}\ \varphi_1 and \mathsf{im}\ \varphi_2 are conjugate subgroups in \mathsf{Aut}(H). If K is infinite, assume that \varphi_1 and \varphi_2 are injective. Prove that H \rtimes_{\varphi_1} K \cong H \rtimes_{\varphi_2} K by constructing an explicit isomorphism. [Hint: Suppose \sigma \varphi_1[K] \sigma^{-1} = \varphi_2[K]. Show that for some a \in \mathbb{Z}, \sigma \varphi_1(k) \sigma^{-1} = \varphi_2(k)^a for all k \in K. Show that the map \psi : H \rtimes_{\varphi_1} K \rightarrow H \rtimes_{\varphi_2} K given by \psi((h,k)) = (\sigma(h), k^a) is a homomorphism. Show that \psi is bijective by constructing a two-sided inverse.]


Let K = \langle x \rangle.

Now \sigma \varphi_1(x) \sigma^{-1} \in \varphi_2[K], so that \sigma \varphi_1(x) \sigma^{-1} = \varphi_2(y) for some y; since K is cyclic, we have y = x^a for some a. Now let k = x^b be arbitrary in K. Then \sigma \varphi_1(k) \sigma^{-1} = \sigma \varphi_1(x)^b \sigma^{-1} = (\sigma \varphi_1(x) \sigma^{-1})^b = \varphi_2(x)^{ab} = \varphi_2(k)^a. Since k is arbitrary, we have \sigma \varphi_1(k) \sigma^{-1} = \sigma_2(k)^a for all k \in K.

We now show that \psi is a homomorphism.

Let (h_1,k_1),(h_2,k_2) \in H \rtimes_{\varphi_1} K. Then

\psi((h_1,k_1)(h_2,k_2)) = \psi((h_1 \varphi_1(k_1)(h_2), k_1k_2))
= (\sigma(h_1 \varphi_1(k_1)(h_2)), (k_1k_2)^a)
= (\sigma(h_1) \sigma(\varphi_1(k_1)(h_2)), k_1^a k_2^a)
= (\sigma(h_1) (\sigma \circ \varphi_1(k_1))(h_2), k_1^a k_2^a)
= (\sigma(h_1) (\varphi_2(k_1)^a \circ \sigma)(h_2), k_1^a k_2^a)
= (\sigma(h_1) \varphi_2(k_1^a)(\sigma(h_2)), k_1^a k_2^a)
= (\sigma(h_1), k_1^a)(\sigma(h_2), k_2^a)
= \psi((h_1,k_1)) \psi((h_2,k_2)).

Thus \psi is a homomorphism.

We now show that \psi is bijective.

  1. Suppose K is infinite, so that \varphi_1 and \varphi_2 are injective. Just as \sigma \varphi_1(k) \sigma^{-1} = \varphi_2(k)^a for all k, there exists b such that \sigma^{-1} \varphi_2(k) \sigma = \varphi_1(k)^b for all k. Combining these results, we see that \varphi_2(k) = \varphi_2(k^{ab}). Since \varphi_2 is injective, k^{1-ab} = 1, and since K is infinite and k arbitrary, we have ab = 1. Thus a = b \in \{1,-1\}.

    Define \chi : H \rtimes_{\varphi_2} K \rightarrow H \rtimes_{\varphi_1} K by \chi((h,k)) = (\sigma^{-1}(h), k^a). Then (\chi \circ \psi)((h,k)) = \chi(\psi((h,k))) = \chi(\sigma(h), k^a) = ((\sigma^{1-}\sigma)(h), k^{aa}) = (h,k), so that \chi \circ \psi = 1. Similarly, \psi \circ \chi = 1. Thus \psi is bijective, and we have H \rtimes_{\varphi_2} K \cong H \rtimes_{\varphi_1} K.

  2. Before proceeding in the finite case, we prove the following lemma, due to Luís Finotti.

    Lemma: Let a,m,n \in \mathbb{Z} such that m|n and \mathsf{gcd}(a,m) = 1. Then there exists \overline{a} \in \mathbb{Z} such that \overline{a} \equiv a mod m and \mathsf{gcd}(\overline{a},n) = 1. Proof: Let d = \mathsf{gcd}(a,n) and write n = mq. Now \mathsf{gcd}(d,m) = 1, so that d|q; we also write a = a^\prime d and q = q^\prime d. Let t be the product of all prime divisors of q^\prime which do not divide d. Finally, let \overline{a} = a + tm. Suppose p is a prime divisor of n. There are three cases:

    1. If p|m, then p \not| a since a and m are relatively prime. Thus p does not divide a + tm = \overline{a}.
    2. If p \not|m and p | q^\prime, we have two cases.
      1. If p|d, then p \not| t by definition. Thus p \not| tm. Also, since p|d, we have p|a. Thus p \not| a + tm \overline{a}.
      2. If p \not| d, then p|t and p \not| a. Thus p \not| a + tm = \overline{a}.
    3. If p \not| m,q^\prime, then since n = mq^\prime d, p | d. Now p | a and p \not| t, so p \not| a + tm = \overline{a}.

    Since no prime divisor of n divides \overline{a}, \mathsf{gcd}(\overline{a},n) = 1. \square

    Now to the main result.

    Suppose K \cong Z_n is finite. Now \mathsf{im}\ \varphi_1 is cyclic of order m where m|n, by Lagrange. Since x generates K, \varphi_1(x) generates \varphi_1[K]. Since conjugation by \sigma is an isomorphism \varphi_1[K] \rightarrow \varphi_2[K], \varphi_2(x)^a generates \varphi_2[K]. Thus \mathsf{gcd}(a,m) = 1. By the Lemma, there exists \overline{a} such that \overline{a} \equiv a mod m and \mathsf{gcd}(\overline{a},n) = 1. Moreover, there exists b such that \overline{a}b \equiv 1 mod n.

    Define \chi : H \rtimes_{\varphi_2} K \rightarrow H \rtimes_{\varphi_1} K by \chi((h,k)) = (\sigma^{-1}(h), k^b). This map is clearly a two sided inverse of \psi; hence H \rtimes_{\varphi_2} K \cong H \rtimes_{\varphi_1} K.

Every group is noncanonically isomorphic to its dual

For any group G define the dual group G (denoted \check{G}) to be the set of all homomorphisms into the multiplicative group of roots of unity in \mathbb{C}. Define a group operation in \check{G} by pointwise multiplication of functions.

  1. Show that this operation on \check{G} makes this set an abelian group. [Hint: show that the trivial homomorphism is the identity and that \chi^{-1}(g) = \chi(g)^{-1}.]
  2. If G is a finite abelian group, prove that \check{G} \cong G. [Hint: write G as \langle x_1 \rangle \times \cdots \langle x_t \rangle and if |x_i| = n_i, define \chi_i to be the homomorphism which sends x_i to e^{2 \pi i/ n_i} and all other x_j to 1. Then show that \chi_i has order n_i and that \check{G} \cong \langle \chi_1 \rangle \times \cdots \langle \chi_t \rangle.]

This result is often stated as follows: a finite abelian group is self-dual. It implies that the subgroup lattice diagram of a finite abelian group is the same when turned upside down. Note however that there is no natural isomorphism G \rightarrow \check{G}; the isomorphism depends on the choice of generators x_i. This is often stated in the form: a finite abelian group is noncanonically isomorphic to its dual.


  1. Let G be a group and A an abelian group, and denote by H = \mathsf{Hom}(G,A) the set of group homomorphisms G \rightarrow A. We define a product on H pointwise. It is clear that this product is associative. Note that the trivial homomorphism 1 : x \mapsto 1 satisfies (1 \cdot \psi)(x) = 1(x) \psi(x) = \psi(x) and (\psi \cdot 1)(x) = \psi(x) 1(x) = \psi(x), so that 1 is an identity under pointwise multiplication. Now let \psi \in H and define \theta : G \rightarrow A by \theta(x) = \psi(x)^{-1}. Note that \theta(xy) = \psi(xy)^{-1} = \psi(x)^{-1} \psi(y)^{-1} = \theta(x) \theta(y), since A is abelian. So \theta \in H. Moreover, (\theta \cdot \psi)(x) = \psi(x)^{-1} \psi(x) = 1 = 1(x) for all x, so that \theta = \psi^{-1}. Hence H is a group. Finally, note that (\psi \cdot \chi)(x) = \psi(x) \chi(x) = \chi(x) \psi(x) = (\chi \cdot \psi)(x), so that H is abelian.
  2. Define each \chi_i as in the problem statement. Since (by this previous exercise) each \chi_i is a homomorphism, \chi_i \chi_j = \chi_j \chi_i for all 0 \leq i,j \leq t.

    Now let (x_i^{a_i}) \in G. We have \chi_i^{n_i}((x_i^{a_i})) = \prod_{j=1}^t (\chi^{n_i}(x_i))^{a_i} = ((e^{2 \pi i/ n_i})^{n_i})^{a_i} = (e^{\pi i})^{2a_i} = 1^{2a_i} = 1. Thus \chi_i^{n_i} = 1. By this previous exercise, there exists a unique group homomorphism \zeta : G \rightarrow \check{G} such that \zeta(x_i) = \chi_i.

    Note that each homomorphism \chi \in \check{G} is uniquely determined by its values at the x_i (by this previous exercise), and that \chi(x_i) is a n_ith root of unity. Thus \chi(x_i) = \chi_i(x_i)^{a_i} for some a_i (since \chi_i(x_i) is a “primitive n_ith root of unity”, whatever that means.). Consider now \prod x_i^{a_i} \in G; note that \zeta(\prod x_i^{a_i})(x_j) = (\prod \zeta(x_i)^a_i)(x_j) since \zeta is a homomorphism, and that this equals \prod (\zeta(x_i)^{a_i}(x_j)) = \prod \chi_i^{a_i}(x_j) = \chi_j^{a_j}(x_j) = \chi(x_j). By uniqueness, then, and because j is arbitrary, \chi = \zeta(\prod x_i^{a_i}). Thus \zeta is surjective.

    Suppose now that \zeta(\prod x_i^{a_i}) = 1. Then by definition \prod \chi_i^{a_i} = 1. Now let 1 \leq j \leq t; we have 1 = \prod \chi_i^{a_i}(x_j) = e^{2 \pi i a_j/ n_j}. Thus n_j divides a_j, hence a_j \equiv 0 mod n_j. Thus we have \prod x_i^{a_i} = 1, so the kernel of \zeta is trivial and thus \zeta is injective.

    Hence G \cong \check{G}.

A quotient by a product is isomorphic to the product of quotients

Let G = A_1 \times A_2 \times \cdots \times A_n and for each i let B_i \leq A_i be a normal subgroup. Prove that B_1 \times \cdots \times B_n \leq G is normal and that (A_1 \times \cdots \times A_n)/(B_1 \times \cdots \times B_n) \cong (A_1/B_1) \times \cdots \times (A_n/B_n).


We begin with some lemmas.

Lemma 1: Let \varphi_1 : G_1 \rightarrow H_1 and \varphi_2 : G_2 \rightarrow H_2 be group homomorphisms. Then \mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). Proof: (\subseteq) Let (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). Then (1,1) = (\varphi_1 \times \varphi_2)(a,b) = (\varphi_1(a), \varphi_2(b)), hence \varphi_1(a) = 1 and \varphi_2(b) = 1. Thus a \in \mathsf{ker}\ \varphi_1 and b \in \mathsf{ker}\ \varphi_2, and (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). (\supseteq) If (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2), then (\varphi_1 \times \varphi_2)(a,b) = (1,1); hence (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). \square

Lemma 2: Let A_1 and A_2 be groups with normal subgroups B_1 \leq A_1 and B_2 \leq A_2. Then B_1 \times B_2 is normal in A_1 \times A_2 and (A_1 \times A_2)/(B_1 \times B_2) \cong (A_1/B_1) \times (A_2/B_2). Proof: Let \pi_1 : A_1 \rightarrow A_1/B_1 and \pi_2 : A_2 \rightarrow A_2/B_2 denote the natural projections. These mappings are surjective, so that \pi_1 \times \pi_2 is a surjective homomorphism A_1 \times A_2 \rightarrow (A_1/B_1) \times (A_2/B_2). By Lemma 1, we have \mathsf{ker}(\pi_1 \times \pi_2) = B_1 \times B_2. The conclusion follows by the First Isomorphism Theorem. \square

The main result now follows by induction on n.