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Let $G$ be a group of order 12 and suppose $G$ does not have normal subgroups of order $4$. How to show that $G$ has a subgroup of order $6$?

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Do you have the Sylow theorems available to you? – Brandon Carter May 13 '11 at 19:58
up vote 2 down vote accepted

An alternative way to finish Peter Patzt's argument is as follows: Once you know that $G$ has a normal subgroup $N$ of order $3$, note that $G/N$ is of order $4$ and therefore abelian. Since normal subgroups of the quotient correspond to normal subgroups of $G$ that contain $N$, and every group of order $4$ contains a normal subgroup of order $2$, this normal subgroup of $G/N$ of order $2$ lifts to a normal subgroup $K$ of $G$ that contains $N$ with $[K:N]=2$. Since $|N|=3$, that means that $K=6$, so you get a normal subgroup of order $6$.

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Thanks! very clear. I accepted this one, because I'm not familiar (yet) with semidirect products. – user6495 May 14 '11 at 16:32
@user6495: Your acceptance of my answeer really does not make much sense. Most of the hard work was done in Peter's answer, and all I did was to suggest an alternative argument for his last three lines. My answer would not make sense without his answer, so I really don't think my answer should be the "accepted" one. – Arturo Magidin May 14 '11 at 19:35

By Sylow's Theorem, the number of Sylow 2-subgroups $n_2 = |G:N_G(P)| \equiv 1 \pmod 2$ where $P$ is any Sylow 2-subgroup of G and NG denotes the normalizer. Since $P$ cannot be normal in $G$, $|G:N_G(P)|\neq 1$ and must be $3$ (the only other odd divisor of $12$). Assuming the Sylow 3-subgroup is not normal either, results to a number of $4$ subgroups of order $4$ by the same argument. This would add up to one neutral element, 8 elements of order $3$, $3$ elements of one Sylow 2-subgroup and at least another element for another Sylow 2-subgroup,totaling at least 13 group elements of $G$.

Hence $G$ has a normal subgroup $N$ of order $3$. $G$ also has a subgroup $H$ of order $2$ contributed to Cauchy (also part of Sylow's Theorem). Now since $N$ is normal in $G$ and $N\cap H = 1$, we have a semidirect product $NH \le G$ of order $6$.

Added after Steve D's comment:

Let $N\triangleleft G$ with $|N|=3$ and $N = \langle\sigma\rangle$. Let $H\le G$ be a subgroup of order $4$. Assume first that $H$ is cyclic and $\tau\in H$ of order $4$. Then $\sigma^\tau \in \{\sigma, \sigma^2\}$, hence $\sigma^{\tau^2}= \sigma$ for the involution $\tau^2$. If $H$ on the other hand is isomorphic to Klein's four-group, and two element $\tau,\tau'\in H$ of order $2$ both map $\sigma^\tau = \sigma^{\tau'} = \sigma^2$ its product, the third involution $\tau\tau'$ maps $\sigma^{\tau\tau'} = \sigma$.

In both cases there is an element $\rho \in H\le G$ of order $2$ that commutes with $\sigma$, hence the semidirect product is direct and cyclic of order $6$.

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You can say more, namely that there must be an element of order 6 (i.e., a cyclic subgroup or order 6). – user641 May 13 '11 at 22:47
thank you very much. – user6495 May 14 '11 at 16:32
You should use \langle and \rangle instead of < and >. – Arturo Magidin May 14 '11 at 19:35
Just a small comment: you don't need to worry about the structure of the Sylow 2-group to prove what I said above: it follows from the normalizer-centralizer theorem since $|Aut(N)|=2$. – user641 May 15 '11 at 7:24

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