# Explicit descriptions of groups of order 45

I know that there are two groups of order 45, and obviously one of them (up to isomorphism) is $\mathbb{Z}_{45}$. I'm trying to understand explicitly what the structure of the other is like.

By Cauchy's Theorem and Sylow's First Theorem, it has a subgroup of order 3, one of order 5, and one of order 9. What this second group of order 45 doesn't have, unlike $\mathbb{Z}_{45}$, is a subgroup of order 15. This rules out an element of order 15, which would generate a cyclic subgroup. I know the group is abelian since we can write it as a product of normal subgroups. Otherwise I'm unsure of what consequences this has, and what the structure of the non-cyclic group of order 45 is.

So basically my question is...how do we obtain an explicit description of the the non-cyclic group of order 45?

By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.

So the group has a single subgroup of order $5$, which must be isomorphic to $\mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $\mathbb{Z}_9$ or to $\mathbb{Z}_3\times\mathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).

Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $N\cap M=\{1\}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And $$|NM| = \frac{|N|\,|M|}{|N\cap M|} = 9\times 5 = 45 = |G|,$$ so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $m\in M$ and $n\in N$. So $G\cong N\times M\cong \mathbb{Z}_5\times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.

Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.

• You wrote $3 \times 5 = 45$. Lol Feb 12 '12 at 5:31
• @Patrick: And you wrote $\mathbb{Z}_3\times\mathbb{Z}_3\mathbb{Z}_5$... Fixed. Feb 12 '12 at 5:33
• Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that. Feb 12 '12 at 5:34
• I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks. Feb 12 '12 at 13:15

If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $\mathbb Z_3$, which you have twice, and $\mathbb Z_5$, which you have once. If you try to "put them all together" you get $\mathbb Z_{45}$, which we clearly expected. Now other combinations would be $\mathbb Z_9 \times \mathbb Z_5$, $\mathbb Z_3 \times \mathbb Z_3\times \mathbb Z_5$, and $\mathbb Z_{15} \times \mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $\mathbb Z_{45}$.

For the other two, since $g.c.d(3,5) = 1$, $\mathbb Z_3 \times \mathbb Z_5$ is isomorphic to $\mathbb Z_{15}$ so we don't have anything new there. But in $\mathbb Z_{15} \times \mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $\mathbb Z_{45}$.

Hope that helps,

• Heh you were quicker than me on that one! Feb 12 '12 at 5:33
• Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks. Feb 12 '12 at 20:22

It's $\mathbb{Z}_{15} \times \mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.