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For a recent qualifier problem, I was to show that if a group $G$ has two solvable subgroups $H,K$ such that $HK=G$ and $H$ is normal, then $G$ is solvable. This is simply a matter of showing that $G/H$ is solvable, and I think is not too difficult. The next part of the question was to find an example of a group with all of the above conditions except the normality condition (i.e. for any two such solvable subgroups, neither is normal in $G$) and show that $G$ is no longer solvable. Does anyone know of a good example? I don't even know that many groups which aren't solvable. I have been told $A_5$ is not solvable, but that is quite a large group, and it seems like it would take a long time to show this in 20 minutes (the time I would have if I was doing a qualifier) if it is even true for $A_5$. I'd love to know what group to look at, so I can prove it myself.

Thanks!

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$A_5$ may appear large, but it is in fact the smallest non-solvable group... –  Mariano Suárez-Alvarez Apr 20 '11 at 16:20
    
So I see, (just read that any group of order less than 60 is solvable). My current strategy consists of showing that $S_5$ is the product of two cyclic groups. –  Jon Beardsley Apr 20 '11 at 17:13

1 Answer 1

up vote 2 down vote accepted

Here's a hint - I don't know what you already know, so if you don't understand, just ask for clarification!

OK, so we're looking for two solvable subgroups $H$ and $K$ of a non-solvable group $G$, such that $HK = G$. The smallest non-solvable subgroup is indeed $A_5$, and every smaller group is solvable. In particular, $A_4$ and $C_5$ are solvable.

Can you find two subgroups of $A_5$, one isomorphic to $A_4$ and another isomorphic to $C_5$ which together generate $A_5$? Can you then show that not only $\langle H,K \rangle = G$ but also $HK = G$?

EDIT: In the previous wrong version I was hinting towards two copies of $A_4$ which together generate $A_5$, but do not have $HK = G$.

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I don't think HK=G in your A4*A4=A5 example. A5=A4*C5 though. –  Jack Schmidt Apr 20 '11 at 17:36
    
@Jack You're completely right! Should've checked before leaving out the right answer. –  yatima2975 Apr 20 '11 at 18:18
    
Thanks @Jack and @Yatima. This is very helpful. –  Jon Beardsley Apr 21 '11 at 15:25

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