5

Since $3\cdot 3\cdot 2\cdot 5=90$, we know that we have a $3$-Sylow subgroup $P_3$ of order $9$, a $2$-sylow subgroup $P_2$ of order $2$, a $ 5$-Sylow subgroup $P_5$ of order $5$.

I know that $P_5 \cong Z_5$ and $P_2 \cong Z_2$, right? But I'm not sure what $P_3$ is isomorphic to, because we cannot necessarily conclude that it is cyclic...since it might have $4$ different elements of order $3$.

So when I'm looking at the different cases for the semidirect products (for example if I look at the case when all of the sylow subgroups are normal), I will just say $G \cong P_3 \times Z_{10}$, right?

I am just asking to make sure if I'm doing it correctly (for this specific case).

Thanks in advance

Watson
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    Is this a problem from a book? Or just one you thought of? Also, you know the $3$-subgroup is abelian, since it has order $p^2$, so it is one of two possible groups. – Potato Apr 13 '13 at 22:19
  • @Potato I didn't make up the question but I made up the solution...anyways, thanks for helping. –  Apr 13 '13 at 22:34
  • Where is the question from? – Potato Apr 13 '13 at 22:58
  • Thanks, Potato. I do not know what I was thinking...Last thing I want to do is confuse!! – amWhy Apr 13 '13 at 23:01
  • @amWhy No problem. This seems quite difficult for a textbook problem. There are $10$ groups of order $90$ up to isomorphism. – Potato Apr 13 '13 at 23:02
  • @Potato I don't know, I think my professor made it up. –  Apr 13 '13 at 23:02
  • @Artus What is the exact homework question? Is it really "Classify all groups of order 90"? – Potato Apr 13 '13 at 23:04
  • @Potato Yes that's the question, but it also says that every group of order 90 has a normal 5-sylow subgroup. So when I was looking at the cases that I need to consider, there are only three cases $n_3,n_2,n_5 =1$; $n_5=1,n_3=1,n_2=5$, and $n_5=1,n_2=5,n_3=10$, right? You already helped me with the first case, so now I have to look at the other 2 cases. –  Apr 13 '13 at 23:10

2 Answers2

5

Here are some hints to help along:

  1. There is a normal subgroup of index 2 (use the regular representation, and look for an odd permutation).
  2. All groups of order 45 are abelian (use Sylow's theorems, and the fact groups of order $p^2$ are abelian).
  3. Your group is a semidirect product of a group of order 45 and one of order 2.
  4. Consider the possible order-2 actions on an abelian group of order 45 (call it H). There are 2⋅2 possible actions when H is cyclic, and 2⋅3 when H is noncyclic. [To see this, decompose $H$ as a direct product, remembering it is abelian.]

Thus there are 4+6=10 possible groups. It is fairly easy to show all are distinct.

3

Some ideas:

Let $\,G\,$ be a group with $\;|G|=90=2\cdot3^2\cdot5\;$ , $\,n_p:=\,$ be the number of Sylow $\,p$-subgroups of $\,G\,$. We denote by $\,P_p, Q_p, R_p\,$ , etc., the different Sylow $\,p$-subgroups:

$$[G:N_G(P_3)]=n_3\in\{1\,,\,10\}$$

Denote $\,M:=P_3\cap Q_3\,$ , and since $\,|M|=1\,,\,3\;$ (why can't this be $\,9\,$ ?) , we get

$$|P_3Q_3|=\frac{|P_3|\,|Q_3|}{|M|}=\frac{81}{|M|}\ge\frac{81}{3}=27$$

so by Lagrange's theorem, $\,|\langle P_3\,,\,Q_3\rangle|=45\,,\,90\,$ (why cannot this be $\,30\,$?) :

$$(i)\;\;\;\;\;|\langle P_3\,,\,Q_3\rangle|=45\implies [G:\langle P_3\,,\,Q_3\rangle]=2\implies \langle P_3\,,\,Q_3\rangle\lhd G$$

$$(ii)\;\;\;\;\;|\langle P_3\,,\,Q_3\rangle|=90\implies M\lhd P_3\,,\,Q_3 \;\;(\text{why?})\implies M\lhd\langle P_3,Q_3\rangle=G$$

So we already have that a group of order $\,90\,$ cannot be simple, but not only that: both $\,P_3\,,\,Q_3\lhd\langle P_3\,,\,Q_3\rangle\,$ and from here we get that in case (i) there's one single Sylow $\,3$-subgroup which is then normal.

Fill in details and try to take it from here at least for some cases.

DonAntonio
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  • @Thanks a lot. There are just two things that are kind of confusing me. The fact that $P_3$ and $Q_3$ are subgroups does not necessarily imply that $P_3Q_3$ is a subgroup as well, right? I also have a question about the notation. What do you mean when you write $<P_3, Q_3>$ (in terms of notation)? –  May 04 '13 at 23:06
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    Precisely because not necessarily $,P_3Q_3,$ is a subgroup is that we take the subgroup generated by them, namely $,\langle P_3,Q_3\rangle,$ – DonAntonio May 04 '13 at 23:08