2
$\begingroup$

Consider the product of cyclic groups $G = C_4 \times C_2$. We see that $H = C_2 \times \{ 0 \}$ is normal in $G$ since $G$ is abelian. The cosets are $H = \{(0, 0), (1, 0)\}$, $(0, 1)H = \{(0,1), (1, 1)\}$, $(2, 0)H = \{(2, 0), (3, 0)\}$, and $(2, 1)H = \{(2, 1), (3, 1) \}$. All three nonidentity cosets have order $2$, so $G/H \cong C_2 \times C_2$. On the other hand, $H \cong J = \{0\} \times C_2 \triangleleft G$, yet $G/J = \{ J, (1, 0)J, (2, 0)J, (3, 0)J \} \cong C_4$. (There's some notation abuse: the $C_2$ and $\{0\}$ in $J = \{0\} \times C_2$ are different from the $C_2$ and $\{0\}$ in $H = C_2 \times \{0\}$.) Therefore, two normal subgroups of $G$ may be isomorphic despite having nonisomorphic quotients.

It's tempting to say the following: $$(C_4 \times C_2)/(C_2 \times \{0\}) \cong C_4/C_2 \times C_2 /\{0\} $$ and $$(C_4 \times C_2)/(\{0\} \times C_2) \cong C_4/\{0\} \times C_2/C_2 \text.$$ But does this trick actually work in general? In other words, is the proposition below true?

Let $G$ be an external direct product $G_1 \times G_2 \times \dots \times G_s$ for some groups $G_1, G_2, \dots, G_s$. Let $H = H_1 \times H_2 \times \dots \times H_s$ be a normal subgroup of $G$ with $H_i \triangleleft G_i$ for each $i = 1, 2, \dots, s$. Then $$ G/H = (G_1 \times G_2 \times \dots \times G_s)/(H_1 \times H_2 \times \dots \times H_s) \cong (G_1/H_1) \times (G_2/H_2) \times \dots \times (G_s/H_s) \text.$$

This result would be like ordinary division of real numbers: $(a \cdot b)/(c \cdot d) = (a/c) \cdot (b/d)$.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, this is true. The proof isn’t bad using the fundamental hom theorem. $\endgroup$ – Randall Oct 16 '20 at 17:56
2
$\begingroup$

There are natural homomorphisms $G_i\to G_i/H_i$, which gives rise to the homomorphism $f:\prod G_i\to \prod(G_i/H_i)$. This is clearly surjective, since each $G_i\to G_i/H_i$ is surjective. We wish to know the kernel.

Let $(g_i)\in\prod G_i$ be such that $f((g_i))=0$. This is equivalent to, for each $i$, $g_i\in\ker(G_i\to G_i/H_i)=H_i$. This means that $\ker f=\prod H_i$.

Thus, by the isomorphism theorem, $\prod G_i/\prod H_i\cong\prod(G_i/H_i)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.