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Let $G$ be a non abelian group of order $36.$ Show that there is more than one sylow $2$-subgroups or more than one sylow $3$-subgroups.

$|G|=2^23^2.$ If $G$ has a unique Sylow $2$-subgroup $H$ then $H\lhd G$ and $O(G/H)=3^2.$ So $G/H$ is abelian. I don't know what to do next?

Please help me !!

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    $\begingroup$ Is this true? (The 3-subgroup in $D_6 \times C_6$ seems to be normal, unless I'm miscalculating. It consists of all the elements of order 3 along with the identity, so can't have non-trivial conjugates.) $\endgroup$ – Billy Jun 15 '13 at 2:56
  • $\begingroup$ You were right. Edited. $\endgroup$ – Sriti Mallick Jun 15 '13 at 8:54
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Suppose that $G$ has only one Sylow $2$-subgroup $H$ and one Sylow $3$-subgroup $K$. Then $H \cap K=\{1\}$, $H$ and $K$ are normal in $G$ and $HK=G$ by cardinality, hence $G \simeq H \times K$. But $H$ and $K$ are abelian, so $G$ too.

Another possibility is to notice that $H$ and $K$ are abelian, and for any $a \in H$, $b \in K$:$$\begin{array}{ll} [a,b] & = a(bab^{-1}) \in H \\ & =(aba^{-1})b^{-1} \in K \end{array}$$hence $[a,b] \in H \cap K =\{1\}$, ie. $a$ and $b$ commute.

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  • $\begingroup$ Sorry there was a typo. I've edited my question. $\endgroup$ – Sriti Mallick Jun 15 '13 at 8:51
  • $\begingroup$ @SritiMallick: So I answered your question. $\endgroup$ – Seirios Jun 15 '13 at 9:04

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