Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm doing exercises from Hungerford's book "Abstract Algebra: An Introduction". The exercise is in section 8.2, numbered 22.

I would like someone to check my proof, as I have reasonable doubts that it is somehow incorrect, mainly because I know that $p$-groups of same order are not necessarily isomorphic. The proof is by induction as you will see, and I'm especially shaky on the inductive step.

Let $G$ and $H$ be finite abelian groups with this property: For each positive integer $m$ the number of elements of order $m$ in G is the same as the number of elements of order $m$ in H. Prove that $G \simeq H$.

Proof: It follows from the problem description that the orders of $G$ and $H$ are equal. It also follows that $p$-groups of both group $G$ and $H$, for every possible $p$, have the same order and the same amount of elements of any order $m$ . Also, number of $p$-groups of both groups is the same. My goal will be to prove that in this case the $p$-groups of same order are isomorphic. For this I will use Lemma 8.6 from Hungerford's book:

LEMMA 8.6 Let $G$ be a finite abelian $p$-group and $a$ an element of maximal order in G. Then there is a subgroup of $K$ such that $G = \langle a\rangle\oplus K$.

If two $p$-groups are of order 2, then clearly they are isomorphic. We map the identity element to identity in the other group, and the second element must have order two which we also map to the element of order 2 in the other group. Assume inductively that all $p$-groups of order less than $n$, with the property from the exercise, are isomorphic to each other. Assume $A$ and $B$, $p$-groups, have order $n$, and have the property from the exercise. Then by the presented lemma, for some elements $a \in G$, $b \in H$ of maximal order, $A = \langle a\rangle \oplus K$ and $B = \langle b\rangle \oplus L$, for some subgroups $K,L$ which have orders less than $A$ and $B$. $K$ and $L$ are of course also $p$-groups, as they are subgroups of $p$-groups. It's easy to see that $K,L$ have the same order, as $|A| = |\langle a \rangle||K|$, $|B| = |\langle b \rangle||L|$, and it follows from the exercise description that the order of highest element in $A$ and $B$, namely $a$ and $b$ are equal. So then, orders of $K$ and $L$ are equal, and by induction hypothesis are isomorphic. $\langle a \rangle$ and $\langle b \rangle$ are isomorphic since both are cyclic groups of the same order. So we have that $A \simeq B$, all $p$-groups of $G$ and $H$ are isomorphic, so $G$ and $H$ are isomorphic.

share|cite|improve this question
Looks good to me. Alternatively, write down the cyclic decomposition of a general abelian $p$-group and express the number of elements of each order in terms of the exponents in the cyclic decomposition. You will see that former determine the latter. – Alex B. Aug 17 '11 at 13:41
ok for the proof. – Louis La Brocante Aug 17 '11 at 13:50
@Barre: I might suggest that when dealing with finite $p$-groups, which must perforce be of order $p^n$ for some $n$, it makes more sense to do induction on the exponent than on the order. That is, the base of your induction ought to be "groups of order $p$", then the inductive hypothesis that it holds for groups of order $p^n$, and then prove it for groups of order $p^{n+1}$. – Arturo Magidin Aug 17 '11 at 14:37
@Arturo thanks for the tip, indeed it seems more reasonable. I'll keep that in mind while doing rest of the exercises :) Also thanks to everyone else for taking their time and reading/confirming validity of my proof. – Malman Aug 17 '11 at 15:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.