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I know cyclic groups of the same order are always isomorphic, but as far as I'm aware finite abelian groups aren't necessarily cyclic. So is this statement true or false, and why?

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    $\begingroup$ The statement is false, for example $C_2$ x$\ C_2$, the Klein-four-group is not isomorphic to $C_4$. $\endgroup$ – Peter Apr 21 '16 at 18:39
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    $\begingroup$ The statement is true iff the order is square free. $\endgroup$ – Arthur Apr 21 '16 at 18:42
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This is not true at all. Simply compare $\Bbb{Z}_2 \times \Bbb{Z}_2$, a non-cyclic group, to $\Bbb{Z}_4$, a cyclic group to see a counterexample.

If you are interested in the structure of finite abelian groups, though, the Fundamental Theorem of Abelian Groups might interest you. Basically, it says that all abelian groups are the direct product of cyclic groups. This means that how the group is structured relies on how the group splits the prime powers up, so if $4$ is split up into $2$ and $2$ like in $\Bbb{Z}_2 \times \Bbb{Z}_2$, then it will be different than if it is just kept as $4$ like in $\Bbb{Z}_4$.

However, if all of the primes in the prime factorization of the order are to the power of one like in $30=2 \cdot 3 \cdot 5$, then the group has to split into cyclic groups of those primes like $\Bbb{Z}_2 \times \Bbb{Z}_3 \times \Bbb{Z}_5$, so all groups of order $30$ are isomorphic. The same goes for $21=3*7$ or $35=7*5$. Numbers that are product of distinct primes like this are called square-free because they have no perfect squares as divisors.

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There are already two different abelian groups of order $4$, namely the cyclic group $C_4$, and the non-cyclic group $C_2\times C_2$. So it is not true. In fact, it gets much worse for bigger (squarefree) group orders: if $n=\prod_{i=1}^rp_i^{k_i}$, then the number of distinct abelian groups of order $n$ is given by $$ \prod_{i=1}^rp(k_i), $$ where $p(k)$ denotes the number of partitions of $k$. So, for example, there are $1$ million different abelian groups of order $49,659,789,817,537,838,957,341,175,342,490,000$, see here.

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