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Fraleigh(7ed) Example37.14 No group of order 36 is simple. Such a group $G$ has either $1$ or $4$ subgroups of order $9$. If there is only one such subgroup, it is normal in $G$. If there are four such subgroups, let $H$ and $K$ be two of them. $H \cap K$ must have at least $3$ elements, or $HK$ would have to have $81$ elements, from $|HK|=|H||K|/|H\cap K|$. Thus the normalizer of $H \cap K$ has as order a multiple of $>1$ of $9$ and a divisor of $36$; hence the order must be either $18$ or $36$. If the order is $18$, the normalizer is then of index $2$ and therefore is normal in $G$. If the order is $36$, then $ H \cap K$ is normal in $G$.

I don't understand the highlighted sentence. It must be from that $N(H \cap K) \supset H$ (or $K$), but why $H\cap K$ is normal in $H$ or $K$? I guess it must be from the first Sylow thoerem(below). But the first Sylow theorem seems to state that $H\cap K$ is a normal subgroup of a subgroup of order $9$, not necessariliy $H$ or $K$, maybe other than $H$ and $K$. How can I conclude that $H \cap K$ is a normal subgroup of $H$ or $K$?

First Sylow Theorem Let $G$ be a finite group and let $|G|=p^n m$ where $n\ge1$ and where $p$ does not divide $m$. Then
1. $G$ contains a subgroup of order $p^i$ for each $i$ where $1 \le i \le n$
2. every subgroup $H$ of $G$ of order $p^i$ is a normal subgroup of a subgroup of order $p^{i+1}$ for $1\le i<n$

Edit: It was very easy. By the first Sylow theorem, $H \cap K$ is a normal subgroup of a subgroup of order $9$, not necessarily $H$ or $K$. But it is still true that $N(H \cap K)$ contains a subgroup of order $9$, so $N(H\cap K)$ has as order a multiple of $9$.

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As Steven says, $H$ and $K$ are abelian. It's a good exercise to show that for each prime $p$ there are only two groups (up to isomorphism) of order $p^2$, and that these are both commutative. –  Dylan Moreland Aug 16 '11 at 4:41
    
You could alternatively try and prove the more general result that no finite group with exactly $k$ Sylow $p$-subgroups for some prime $p$ and $k \le 4$ can be nonabelian simple. –  Derek Holt Aug 16 '11 at 9:13
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up vote 7 down vote accepted

Groups of order 9 must be commutative, so any subgroup of $H$ or $K$ is normal relative to it.

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Thanks for giving another point of view. And I revised my mistake and completed the argument using the Sylow theorem. –  Gobi Aug 16 '11 at 4:56
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