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Let $N$ be a normal subgroup of $S_4$.

I have proven that $|N|\ne 2,3,8$.

Yet, I don't know how to prove that $|N|\neq 6$.

Should I compute all subgroups and check this?

user26857
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Number 9
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5 Answers5

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A normal subgroup is a union of conjugacy classes, one of which must be the identity element.

The non-identity conjugacy classes in $S_4$ have sizes $3$, $6$, $6$, and $8$, and no subset of these numbers sums to $5$.

Jim Belk
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    Incidentally, a similar proof shows that $A_5$ is simple. The non-identity conjugacy classes have sizes 12, 12, 15, and 20, and no subset of these numbers sums to one less than a divisor of 60. – Jim Belk Jul 13 '15 at 04:12
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Suppose $S_4$ has a normal subgroup of order $6$ and call it $N$. Then since $|N|=6$, it contains an element of order $2$ and the elements of order $2$ in $S_4$ are either a cycle of length $2$ or product of two cycles each of length $2$.

Case 1. If $N$ contains a cycle of length $2$, then it will contain $\{(1~2),(1~3),(1~4),(2~3),(2~4),(3~4)\}$ because all are conjugate and $N$ is normal. So order of $N$ must be greater than $7$, which is a contradiction.

Case 2. If $N$ contains an element which is a product of two cycles each of length $2$, then it will contain $\{(1~2)(3~4),(1~3)(2~4),(1~4)(2~3)\}$ because all are conjugate and $N$ is normal. So $V_4$ is normal subgroup of $N$. So from Lagrange's theorem, $4\mid6$ which is a contradiction.

user26857
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brinkle martyn
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There are only two groups of order $6$: $\mathbb{Z}/6\mathbb{Z}$ and $S_3$. Since $S_4$ doesn't have any elements of order $6$, it cannot contain a copy of $\mathbb{Z}/6\mathbb{Z}$. Thus if $|N| = 6$, it must be isomorphic to a copy of $S_3$; relabel the elements in $S_4$ s.t. $N = \{e,(12),(13),(23),(123),(132)\}$. However, this isn't normal as $(14)(13)(14) = (34) \notin N$.

Marcus M
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  • How do you know that any copy of $S_3$ in $S_4$ can be "relabeled" to be the standard copy? Asked differently, what does $(14)$ even mean after the "relabeling"? – Jim Belk Jul 13 '15 at 04:18
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    You can always relabel the letters themselves---$1,2,3$ and $4$---but if you want to avoid relabeling, take any transposition outside of $N$. – Marcus M Jul 13 '15 at 04:28
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    But some proof is required that any copy of $S_3$ in $S_4$ is conjugate to the standard copy. This isn't true in $S_5$, for example -- the subgroup generated by $(1,2,3)$ and $(1,2)(4,5)$ is isomorphic to $S_3$. – Jim Belk Jul 13 '15 at 04:36
  • I agree. There is something (fairly straightforward) to be shown. – Marcus M Jul 13 '15 at 12:19
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    I guess the point is that $S_3$ is generated by an element of order $3$ and one of order $2$. Up to conjugacy, the only element of order $3$ in $S_4$ is $(1,2,3)$. Without loss of generality, the only elements of order two are then $(1,2)$, $(1,4)$, and $(1,2)(3,4)$. But one can check that the latter two do not generate a subgroup of order $6$ with $S_3$, so then you're right that the only possibility is $(1,2,3)$ and $(1,2)$, i.e. the "standard" copy of $S_3$. – Jim Belk Jul 13 '15 at 13:43
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Suppose $N$ is normal subgroup of order $6$, then $S_4/N \cong C_4$ or $S_4/N \cong C_2\times C_2$ both are abelian. It follows that $[S_4,S_4]=A_4 \subseteq N$, which leads to contradiction.

user26857
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Chiranjeev_Kumar
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Suppose $N\unlhd S_4$ has order $6$. The action $\varphi$ of $S_4$ by left multiplication on the quotient $S_4/N$ (of order $4$) has $\ker\varphi=$ $\bigcap_{\sigma\in S_4}\sigma N\sigma^{-1}=$ $N$. Therefore, $(S_4\ge)\operatorname {im}\varphi\cong$ $S_4/N\cong C_4$ or $C_2\times C_2$. Since neither $C_4$ nor $C_2\times C_2$ have elements of order $3$, the eight $3$-cycles of $S_4$ must all lie in the kernel$^\dagger$: too many for a kernel of six elements. Contradiction.

$^\dagger$The order of the image of an element under a homomorphism divides the order of the element.

citadel
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