# Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?

I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states

Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.

Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.

After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you.

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If $G$ is abelian, then any subgroup is normal. – Arturo Magidin Sep 16 '11 at 2:49
@Arturo, ah yes of course, I'll remove it since it's redundant. – yunone Sep 16 '11 at 2:50
I think you need to use the theorem on the structure of finite abelian groups. – lhf Sep 16 '11 at 2:54
Referring to a text written by (Serge, presumably) Lang as simply "Lang" leaves a lot of ambiguity. Do you mean Lang's Algebra? Please be more specific! – Pete L. Clark Sep 16 '11 at 2:55
My apologies, I'll be more specific in the future. – yunone Sep 16 '11 at 5:52
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Since a finite abelian group is the direct sum of its $p$-parts, it suffices to establish the result when $G$ is a finite abelian $p$-group.

If $G=C_{p^{a_1}} \oplus\cdots\oplus C_{p^{a_k}}$, with $1\leq a_1\leq\cdots \leq a_k$, and let $Q$ be a quotient of $G$. Then $Q$ is a finite abelian $p$-group that is generated by $k$-elements (the images of the generators of $G$), and so when we express it as a direct sum of cyclic $p$-groups, it will have at most $k$ direct summands, $$Q \cong C_{p^{b_1}}\oplus\cdots\oplus C_{p^{b_m}},$$ $1\leq b_1\leq \cdots\leq b_m$, $m\leq k$.

Now, $b_m\leq a_k$, because every element of $G$ is of order dividing $p^{a_k}$, hence the same is true for $Q$. So $C_{p^{a_k}}$ has a subgroup of order $p^{b_m}$.

Likewise, $b_{m-1}\leq a_{k-1}$ (count the number of elements of order greater than $p^{a_{k-1}}$ in $G$; an element of order greater than $p^{a_{k-1}}$ in $Q$ must be an image of one of these). So you can find a subgroup of $C_{p^{a_{k-1}}}$ of order $p^{b_{m-1}}$.

Continue this way until you get all the cyclic summands you need out of the cyclic summands of $G$ to construct a subgroup isomorphic to $Q$.

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 Thanks Arturo, this answer made the most immediate sense to me at my level. – yunone Sep 16 '11 at 5:47 Pardon me Professor, can you give a more explicit explanation of why $b_{m-1}\leq a_{k-1}$? I don't fully understand the suggested counted argument. Thanks. – Vika Dec 9 '11 at 11:59 @Vika: The image of $p^{a_{k-1}}G$ in $Q$ is $p^{a_{k-1}}Q$; since $p^{a_{k-1}}G$ is cyclic, that means that $p^{a_{k-1}}Q$ is cyclic; so at most one of the $b_i$ is greater than $a_{k-1}$, and that will be (if any) $b_{m}$. – Arturo Magidin Dec 9 '11 at 16:16 Is $p^{a_{k-1}}G$ the set of all elements of $G$ multiplied by $p^{a_{k-1}}$? This sends the first $a_{k-1}$ summands to $0$, so $p^{a_{k-1}}G\cong p^{a_{k-1}}C_{p^{a_k}}$? How does this have image $p^{a_{k-1}}Q$ under the natural projection into $Q$? – Vika Dec 9 '11 at 17:27 @Vika: Given any abelian groups $A$ and $B$, and an onto homomorphism $A\to B$, and for every $n\in \mathbb{N}$, $nA$ maps onto $nB$: given $nb\in nB$, let $a\in A$ such that $a\mapsto b$. Then $na\in nA$ maps to $nb$. Conversely, given any element $b$ in the image of $nA$, there exists $a\in A$ such that $b$ is the image of $na$, hence $b$ is $n$ times the image of $a$. – Arturo Magidin Dec 9 '11 at 17:43
The fact that if $G$ is abelian every subgroup is normal appears on page 1.