If G is an abelian finite group, and H is a subgroup of G, How can I prove that G/H is isomorphic to another subgroup N of G?.
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Welcome to math.SE. Could you show us what you've tried so far, or what's your thoughts about the problem? Thanks! – Jay Zha May 22 '17 at 21:51
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This is false for example when $G$ is cyclic of order $4$ and $|H|=2$. – Derek Holt May 22 '17 at 22:09
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@Derek What was the duplicate you considered? It can be added (edited) into the "pink" banner as such. – amWhy May 23 '17 at 01:27
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@DerekHolt No, it's true. The question doesn't ask for $N\cap H={0}$ – egreg May 23 '17 at 06:03
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This is not a duplicate of the linked question. – egreg May 23 '17 at 06:14
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@egreg It asks for "another subgroup", which means a subgroup other than $H$. The word "another" means different from the original. – Derek Holt May 23 '17 at 08:28
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@DerekHolt Disputable. Anyway there's no mention about splitting. – egreg May 23 '17 at 08:39
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See subgroups of finite abelian groups for references. – Alex Vong May 23 '17 at 10:03
2 Answers
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I'll write abelian groups additively.
You can use the character group: if $G$ is a finite abelian group, then its character group is $\hat{G}=\operatorname{Hom}(G,\mathbb{Q}/\mathbb{Z})$.
From the structure theorem of abelian groups, it follows that $G$ is the direct sum of cyclic $p$-groups. If $C$ is a finite cyclic $p$-group, then $\hat{C}\cong C$, hence there is a (non canonical) isomorphism $\varphi\colon G\to\hat{G}$.
If $H$ is a subgroup of $G$, consider $H'=\varphi(H)$, so $\hat{G}/H'\cong G/H$. Now prove that, defining $$ N=\{x\in G:\chi(x)=0,\text{ for all }\chi\in H'\}, $$ you have $N\cong \hat{G}/H'$.
egreg
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3Really doubtful that the OP has encountered the material necessary to understand your answer. – amWhy May 23 '17 at 01:28
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HINT:
Because $G$ is abelian, every subgroup will be normal. Therefore, it makes sense to consider $G/H$ for a subgroup $H$.
Then, use the first isomorphism theorem such that $H = ker(f)$