Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Classify all groups of order $1805.$ It may help to note that $\left(\begin{array}{c} 0 &-1\\ 1 & 4 \end{array}\right)$ has order $5$ in $GL_2(\mathbb{F}_{19}).$

My idea: Observe that $1805=5^{1}\times 19^{2}.$ So, $G\cong \mathbb{Z}_5\times \mathbb{Z}_{19^2}$ and $G\cong \mathbb{Z}_5\times \mathbb{Z}_{19}\times \mathbb{Z}_{19}.$ I don't understand what the hint is for. Can anyone help me? Thanks.

share|cite|improve this question
Note that the groups you have given are abelian, while the question asks you to classify all groups of order 1805. – Rankeya Jul 16 '12 at 6:36
The hint is intended to help you find a way to construct a group of order 1805 as a semidirect product. Then when you have done that you should consider if you have found them all. – Tobias Kildetoft Jul 16 '12 at 6:49
Do you know about semi direct products? – Dylan Moreland Jul 16 '12 at 6:50
@DylanMoreland No, I was not taught semi direct product when I took this class. (I took the class at a different university than from where I am preparing for prelim.) – Lyapunov Jul 16 '12 at 6:57
I found an answer : – Lyapunov Jul 16 '12 at 7:08
up vote 4 down vote accepted

I see you've found an answer, but I'll just post what I had been writing up anyway.

Let $G$ be a group of order 1805. By applying Sylow's Theorems, we can tell that $G$ must have a normal Sylow 19-subgroup, call it $N$. Then we can write $G = NH$, where $H$ is any subgroup of order 5.

Now $H$ acts on $G$ by conjugation, and since $N$ is normal, this action restricts to an action on $N$. We can check to see what can happen depending on the isomorphism type of $N$. As a group of order $19^2$, $N$ is either $\mathbb Z_{19^2}$ or $\mathbb Z_{19} \times \mathbb Z_{19}$.

If $H$ acts trivially on $N$, then we have either $G \cong \mathbb Z_{5} \times \mathbb Z_{19^2}$ or $G \cong \mathbb Z_{5} \times\mathbb Z_{19} \times \mathbb Z_{19}$, which are the cases you've already found.

Now let's see which kinds of nontrivial actions could be happening.

If $N \cong \mathbb Z_{19^2}$, then $\operatorname{Aut}(N) \cong (\mathbb Z_{19^2})^\times$, which has order $\varphi(19^2) = 342$, which is not divisible by 5, so we can't have a nontrivial action in this case.

If $N \cong \mathbb Z_{19} \times\mathbb Z_{19}$, then $N$ is a 2 dimensional vector space over the field $\mathbb F_{19}$, and so $\operatorname{Aut}(N) \cong \operatorname{Gl}_2(\mathbb F_{19})$, which has order $(19^2-1)(19^2-19) = 2^4 \cdot 3^4 \cdot 5 \cdot 19$. Thus if $H$ has a nontrivial action on $N$, then it acts like a Sylow 5-subgroup of $\operatorname{Aut}(N)$. Since these subgroups are all conjugate and are cyclic, any two nontrivial actions of $H$ on $N$ will give isomorphic groups. This fact can be found on page 184 of Dummit and Foote's Abstract Algebra.

share|cite|improve this answer

Your Answer


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.