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Prove that a group of order 36 must have a normal subgroup of order 3 or 9.

Let n2 be the number of 2-Sylow subgroups of G (with |G|=36). Then n must be 1 or 3. Let n3 be the number of 3-Sylow subgroups of G. then n3=1 or n3=4 if n3=1 we have 1 3-sylow group of order 9. and it is also a normal group(from sylow theorem ) if n2=1 there is normal group of order 4 but I cant show normal group of order 3.

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If not, the 3-Sylows would be core-free, so your group would embed in $S_4$. –  Martino Oct 30 '12 at 17:45
    
See this link math.stackexchange.com/q/57754/8581. –  Babak S. Oct 30 '12 at 17:49

1 Answer 1

up vote 5 down vote accepted

Assume that $G$ has $4$ Sylow $3$-groups, as you noted $(n_3=1+3k|4, n_3\neq1)$. Defining the conjugation action on the set of these $4$ Sylow $3$-groups; we have the induced homomorphism $\phi: G\longrightarrow S_4$. So, $\frac{G}{\ker\phi}\hookrightarrow S_4$, but $3^2\bigg||G|=36$ and dose not divide $|S_4| = 24$ so, $\ker\phi\neq 1$ and of course $\ker\phi\neq G$. This means that there must be non-trivial kernel, which is a non-trivial normal subgroup of G.

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Hello, dear friend! +1 for you ;-) –  amWhy Apr 2 '13 at 1:55
    
@amWhy: Thanks Amy. I have been slow these days cause there have been some jobs around home. Everyday something new. and I an't perform as always. -100. :-( –  Babak S. Apr 2 '13 at 6:21

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