# Q question about proving isomorphism of abelian groups

Suppose that $$\mathbb{Z}_n^{+}$$ denotes the cyclic group of order $$n$$.

Question a: Consider the group $$G=\mathbb{Z}_{n_1}^{+}\times \mathbb{Z}_{n_2}^{+}\times \ldots \mathbb{Z}_{n_k}^{+}$$ where $$n_1,\ldots,n_k$$ are pairwise relatively prime. Let $$H$$ be a finite group such that $$|H|=|G|$$. Is it true that in order to prove that $$H\cong G$$, it is suffices to prove that $$H$$ contains an element $$a_i$$ of order $$n_i$$ for each $$i$$ ? If not, how to prove that $$H\cong G$$ without describing a specific isomorphism?

Question b Consider the group $$G=\mathbb{Z}_{p^{k_1}}^{+}\times \mathbb{Z}_{p^{k_2}}^{+}\times \ldots \mathbb{Z}_{p^{k_r}}^{+}$$ where $$p$$ is a prime number and $$k_1\leq k_2\leq\ldots\leq k_r$$. Let $$H$$ be a finite group such that $$|H|=|G|$$. How to prove that $$H\cong G$$ without describing a specific isomorphism?

Thanks!

• For question a, its sufficient. For question b, are the primes pairwise distinct? May 21, 2019 at 12:16
• I don't think it's sufficient, as he states it, considering that the $a_i$'s might not generate the group. May 21, 2019 at 12:17
• Thanks @Wuestenfux. In Question b, it is the same prime.
– boaz
May 21, 2019 at 12:23
• That assumption is not enough. I could take $H=G\times \mathbb{Z}_{n_{k+1}}$ where $n_{k+1}$ is such that $n_1,\ldots,n_{k+1}$ are still relative primes. I just chose $\mathbb{Z}$ because it's faster to write. May 21, 2019 at 12:34
• I think you have to either assume that the elements that you found generate the group or assume that $H$ is a subgroup. Maybe assuming that they have the same cardinality would also be enough. May 21, 2019 at 12:36

$$\textbf{Question a:}$$ I'm assuming you mean $$\textit{pairwise}$$ relatively prime, since them being just mutually relatively prime wouldn't do much for you, I don't think.

Assuming that, the answer is $$\textbf{yes}$$ and it comes from the fact that, if $$H_i,H_j are the subgroups generated by $$a_i$$ and $$a_j$$, respectively, with $$i\neq j$$, then $$H_i\cap H_j< H_i\implies |H_i\cap H_j|\mid n_i$$ and $$H_i\cap H_j. Therefore $$H_i\cap H_j=\{0\}$$. This implies that the $$a_i$$'s generate $$H$$.

Then you can just consider the homomorphism $$\varphi:G\to H$$ determined by sending $$\mathbb{Z}_{n_i}^+$$ to $$H_i$$ in the obvious way $$(1\mapsto a_i)$$. Since the $$a_i$$'s generate $$H$$, it's surjective and therefore also injective, since the groups are finite with the same cardinality.

$$\textbf{Question b:}$$ Here the assumption that $$|H|=|G|$$ is not enough: If $$G=\mathbb{Z}_2^+\times\mathbb{Z}_4^+$$ and $$H=\mathbb{Z}_8^+$$, your conditions are verified: $$2$$ and $$4$$ have orders $$4$$ and $$2$$, respectively, in $$H$$. However, the groups are not isomorphic. You must further assume that such elements generate $$H$$, which in the previous question came free at the cost of their orders being pairwise relatively prime. The isomorphism is then given the same way.

For question (a), it is not sufficient. You need stronger conditions:

• either that $$n_i$$ are pairwise coprime, i.e., $$\gcd(n_i,n_j)=1$$ instead of just $$\gcd(n_1,n_2,\dots,n_k)=1$$.
• or the elements $$a_i$$ generate the group $$H$$.

To see why merely $$n_1,n_2,\dots,n_k$$ relatively prime fails, consider $$n_1=2^3\times 3^2\times 5$$, $$n_2=3^3\times 5^2\times 7$$, $$n_3=5^3\times 7^2\times 2$$, $$n_4=7^3\times 2^2\times 3$$. Then $$G$$ is isomorphic to $$\prod_{p=2,3,5,7} (C_{p^3}\times C_{p^2}\times C_p)$$ and the group $$H=C_{2^6\times 3^6\times 5^6\times 7^6}$$ has the same number of elements as $$G$$, and since it has an element of order $$2^6\times3^6\times5^6\times7^6$$ it has an element of every factor of $$2^6\times3^6\times5^6\times7^6$$. But $$G$$ is not cyclic, because every element has order divisible by $$2^3\times3^3\times5^3\times7^3$$.

Similarly for question (b), if you can exhibit $$a_i$$ of order $$p^{k_i}$$ in $$H$$ such that the $$a_i$$s generate $$H$$, then $$G\cong H$$. But of course that is pretty much describing a specific isomorphism.