32

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.

Martin Sleziak
  • 53,687
yunone
  • 22,333

2 Answers2

28

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 $Q$.

Arturo Magidin
  • 398,050
  • Thanks Arturo, this answer made the most immediate sense to me at my level. – yunone Sep 16 '11 at 05:47
  • 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 09 '11 at 11:59
  • 1
    @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 09 '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 09 '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 09 '11 at 17:43
  • I understand that now, thanks. Is $p^{a_{k-1}}G$ cyclic since it's isomorphic to some subgroup of $C_{p^{a_k}}$? I then understand then that $p^{a_{k-1}}Q$ is cyclic as the image of a cyclic group. But why does that imply at most one $b_i$ is greater than $a_{k-1}$? – Vika Dec 09 '11 at 17:54
  • @Vika: $p^{a_{k-1}}G$ is cyclic because $p^iC_{p^b} = 0$ if $b\leq i$, and $p^iC_{p^b} \cong C_{p^{b-i}}$ if $i\lt b$. So, as you noted, $p^{a_{k-1}}G \cong p^{a_{k-1}}C_{p^{a_k}} \cong C_{p^{a_k-a_{k-1}}}$ is cyclic. But on the other hand, the same argument applied to $Q$ tells you that $p^{a_{k-1}}Q$ has as many nontrivial cyclic summands as there are $b_i$ that are greater than $a_{k-1}$; since it is in fact cyclic (being a quotient of a cyclic group), there is at most one such cyclic summand, so there is at most one such $b_i$. Lather, rinse, and repeat. – Arturo Magidin Dec 09 '11 at 18:01
  • A very large thank you for your patience. – Vika Dec 09 '11 at 18:13
  • 1
    May I know how you continue the argument to show that $b_{m-2}\leq a_{k-2}$? since $p^{a_{k-2}}G=p^{a_{k-2}}C_{p^{a_{k-1}}}\bigoplus p^{a_{k-2}}C_{p^{a_{k}}}$ is not cyclic anymore. – Wang Kah Lun Aug 05 '19 at 13:02
  • @AlanWang and future ppl wondering: since $p^{a_{k-2}}G$ is generated by 2 elements its image in the quotient must be generated by $\leq 2$ elements and so all summands except $C_{p^{b_{m-1}}} \oplus C_{p^{b_m}}$ must be killed by $p^{a_{k-2}}G$. Indeed the quotient can't have $\geq 3$ summands bc that would mean it needed 3 generators (to see this map each summand mod p to get $(\mathbb Z / p \mathbb Z)^3$ which cannot be generated by 2 elts via linear alg or counting argument). – vacant Feb 09 '22 at 19:54
13

I have some notes on (mostly finite) abelian groups for an undergraduate audience here.

The fact that if $G$ is abelian every subgroup is normal appears on page 1.

The result you are asking about is Theorem 19 on page 8 of my notes. Beware that although a complete proof is in the notes, it takes a little while to get there...the point is that this uses, in addition to the basic character theory of finite abelian groups, the fact that a finite abelian group is non-canonically isomorphic to its character group, which in turn uses the main structure theorem for finite abelian groups.

Added: It is possible to dispense with the character theory (although to my taste this is a nice, clean way to phrase it), but it does not seem possible to avoid the structure theorem, which is a famous, and famously nontrivial, result. Note that in particular Arturo's nice answer does not use character theory but does use the structure theorem...twice.

Martin Sleziak
  • 53,687
Pete L. Clark
  • 97,892
  • Indeed; I don't think you would be able to dispense with the structure theorem, given that the statement is so general. If you have no handle on what the quotient can look like, it seems that it would be rather hard to come up with a proof. – Arturo Magidin Sep 16 '11 at 03:16
  • Thanks Pete, I've actually been reading through a few of the algebra, number theory, and topology handouts you have on your website. The ones that are accessible to me are really quite clear and helpful. – yunone Sep 16 '11 at 05:51