Q question about proving isomorphism of abelian groups

Question

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!

Answer 1

$\textbf{Question a:}$ 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 $H_i,H_j<H$ 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 $H_i\cap H_j<H_j\implies |H_i\cap H_j|\mid n_j$. 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.

Answer 2

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.