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The question asks to find the center of $\mathbb{Z}_5 \times \mathbb{Z}_{15}$. I believe the answer is that the center is the whole group, since $\mathbb{Z}_5 \times \mathbb{Z}_{15} \cong \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_3$, and since the group is of the form $\mathbb{Z}_{p_{1}^{i_1}} \times \cdots \times \mathbb{Z}_{p_{n}^{i_n}}$ where the $p_j$'s are primes (not necessarily distinct), then this must mean it is isomorphic to some Abelian group of order 75. Is this reasoning correct?

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    $\begingroup$ The direct sum of two cyclic groups definitely is abelian and hence the center is the whole group. $\endgroup$ – Mathematician 42 Nov 28 '16 at 21:20

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