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 $n_i$th 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 $n_i$th 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 $n$.