Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 9 down vote accepted

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 $45$.

share|improve this answer
    
You wrote $3 \times 5 = 45$. Lol –  Patrick Da Silva Feb 12 '12 at 5:31
    
@Patrick: And you wrote $\mathbb{Z}_3\times\mathbb{Z}_3\mathbb{Z}_5$... Fixed. –  Arturo Magidin 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. –  Patrick Da Silva 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. –  Alex Petzke 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,

share|improve this answer
    
Heh you were quicker than me on that one! –  Patrick Da Silva 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. –  Alex Petzke Feb 12 '12 at 20:22

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.