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Assume that we have the following commutative diagram with groups and homomorphisms where $b$ and $c$ are injective homomorphisms

$\begin{array}[c]{ccc} A&\stackrel{a}{\rightarrow}&B\\ \downarrow\scriptstyle{b}&&\downarrow\scriptstyle{c}\\ C&\stackrel{d}{\rightarrow}&D \end{array}$

Is it true that $C/b(A) \cong D/c(B)$? Why?

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3 Answers

Hint: Let $A$ and $B$ both be trivial groups. What does your question ask in this case?

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are they some conditions on the groups and the homomorphisms to have correct answer? –  Ronald Jun 1 '13 at 19:00
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The answer is no. Put $A = B = C = G$ and $D = G \times G$. $a$ , $b$ identity , $c$ $d$ inclusion on the first factor.

Then $C/b(A) = \{e\}$ and $D/c(B) \simeq G$

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are they some conditions on the groups and the homomorphisms to have correct answer? –  Ronald Jun 1 '13 at 19:15
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It's not true, take $A=B=C=0$ (the trivial group) and $D$ any nontrivial group.

It is however true that $d$ induces an isomorphism $C/b(A) \to d(C)$. For this isomorphism to be an isomorphism $C/b(A) \simeq D/c(B)$, you would need $B$ to be trivial (since $c$ is injective) and $d$ to be surjective. Injectivity of $b$ doesn't matter.

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are they some conditions on the groups and the homomorphisms to have correct answer? –  Ronald Jun 1 '13 at 19:15
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@Danial: I edited my answer to give a possible condition. –  Najib Idrissi Jun 1 '13 at 19:26
    
thanks a lot :) –  Ronald Jun 1 '13 at 21:19
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