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Suppose $\varphi:G \rightarrow H$ is an isomorphism of groups. $A \le G, B \le H$ and we have the commutative digaram, $$\require{AMScd} \begin{CD} A @>\tilde{\varphi}>> B \\ @Vj_1 V V @VVj_2 V\\ G @>>\varphi > H \end{CD} $$ where $j_1, j_2$ are injective homomorphisms, and $\tilde{\varphi}$ is an isomorphism. Can we show that $|G:j_1(A)|=|H:j_2(B)|$ (the cardinality of the cosets)?

Problem I: Is this proof correct?

Consider the map $[g]_A \mapsto [ \varphi(g)]_B$, mapping cosets of $j_1(A)$ to cosets of $j_2(B)$ under $\varphi$.

The map is well defined: If $[\varphi \circ j_1(a)]_B= [j_2 \circ \tilde{\varphi}(a)]_B = [e]_B$ by commutativity.

The map is injective: $[\phi(g_1)]_B = [\phi(g_2)]_B$ iff $[\phi(g_2^{-1} \cdot g_1)]=[e]_B$. We also have $$\varphi(g) \in j_2(B) \Rightarrow \varphi(g)= j_2 \circ \tilde{\varphi}(a) = \varphi \circ j_1(a) \Rightarrow g=j_1(a) \in j_1(A)$$ as $\varphi$ is injective, and $\tilde{\varphi}$ is surjective. Injectivity follows.

The map is surjective: holds as $\varphi$ is surjective.

Hence $|G:j_1(A)|=|G:j_2(B)|$

Problem II: In showing well defineness we only need commutativity of diagram. In showing injectivity and surjectivity, we only need $\varphi$ to be bijective and $\tilde{\varphi}$ to be surjective. Are all other conditions unnecessary? This seems weird...

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    $\begingroup$ So $j$ is not necessarily the inclusion? then try $A=G=H=\Bbb Z$, $B=2\Bbb Z$, $\tilde \phi$ multiplication by $2$, $j$ on the right divisoin by $2$. $\endgroup$ Commented Dec 12, 2017 at 2:29
  • $\begingroup$ Ok, thanks, - I was actually confused about this statement: math.stackexchange.com/questions/2562599/… , I thought I could simplify it to the groups case... what is happening? $\endgroup$
    – Bryan Shih
    Commented Dec 12, 2017 at 3:09
  • $\begingroup$ @HagenvonEitzen, sorry, I think I have misstated the problem. $\endgroup$
    – Bryan Shih
    Commented Dec 12, 2017 at 3:42

1 Answer 1

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In showing well defineness we only need commutativity of diagram. In showing injectivity and surjectivity, we only need $\varphi$ to be bijective and $\tilde{\varphi}$ to be surjective.

I’ll reprove this claim.

Well defineness. Let $g_1,g_2\in G$ and $g_1j_1(A)=g_2j_1(A)$. Then $g_2^{-1}g_1\in j_1(A)$. Pick $a\in A$ such that $g_2^{-1}g_1=j_1(a)$. Then $\varphi(g_2^{-1}g_1)=\varphi(j_1(a))=j_2\tilde{\varphi}(a)\in j_2(B)$, so $\varphi(g_1) j_2(B)=\varphi(g_2) j_2(B)$.

Injectivity. For each $g\in G$ a preimage of a coset $\varphi(B) j_2(B)$ is a set

$$\{h\in G: \varphi(h) j_2(B)=\varphi(g) j_2(B)\}= \{h\in G: \varphi(g)^{-1} \varphi(h)\in j_2(B)\}= \{h\in G: \varphi(g^{-1}h)\in j_2(B)\}= \{h\in G: g^{-1}h\in \varphi^{-1}(j_2(B))\}.$$

So we have to show that $\varphi^{-1}(j_2(B))=j_1(A)$. Let $g\in j_1(A)$. Pick $a\in A$ such that $g=j_1(a)$. Then $\varphi(g)=\varphi(j_1(a))=j_2\tilde{\varphi}(a)\in j_2(B)$. On the other hand, let $g\in \varphi^{-1}(j_2(B))$. Then there exists an element $b\in b$ such that $\varphi(g)=j_2(b)$. Since the map $\tilde{\varphi}$ is surjective, there exists an element $a\in A$ such that $b=\tilde{\varphi}(a)$. Then $\varphi(j_1(a))=j_2\tilde{\varphi}(a)=j_2(b)= \varphi(g)$. Since the map $\varphi$ is injective, $ j_1(a)=g$, so $g\in j_1(A)$.

Surjectivity. Holds as $\varphi$ is surjective.

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