Show $\Bbb{Z}_{30},\ \Bbb{Z}_5 \times D_3,\ \Bbb{Z}_3 \times D_5$ and $D_{15}$ are not isomorphic where $D_n$ is the dihedral group.

Question

I'm looking for a hint for the following problem:

Show that no two elements of the following list are isomorphic ($D_n$ is the dihedral group of order $2n$). $$\mathbb{Z}_{30},\ \mathbb{Z}_5 \times D_3,\ \mathbb{Z}_3 \times D_5 \text{ and }D_{15}$$

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My attempt

I know that $\mathbb{Z_{30}}$ is abelian and $\mathbb{Z}_a\times D_b$ with $a\in\{3,5\}$ and $b\in\{3,5\}\setminus \{a\}$ isn't (because $D_n$ isn't, for all $n$) and thus $\mathbb{Z}_{30}$ can't be isomorphic to any of the other elements of the list. The question is now reduced to the same problem but now with the list $$\mathbb{Z}_5 \times D_3,\ \mathbb{Z}_3 \times D_5 \text{ and }D_{15}.$$ Here I'm completely stuck, I feel like I'm close to finding a propperty held by $\mathbb{Z}_5 \times D_3$ and not by $\mathbb{Z}_3 \times D_5$ but I'm not quite there. So I come here looking for a hint. Thanks in advance.

Answer 1

It seems easier to me to use the following statement. By the way, it is useful to remember it in any case.

The group $D_n$ for odd $n$ has exactly $n$ involutions, i.e., elements of order $2$.

If, for example, we define $D_n$ as a group of symmetries of a regular $n$-gon, then the elements of order $2$ are exactly all $n$ reflexions.

Now it is easy to see that the number of involutions for our groups is: $3$, $5$, and $15$. So these three groups are pairwise non-isomorphic.