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Question is to prove that :

$S_4$ does not have a normal subgroup of order $8$

I do not have any specific idea how to proceed but:

Assuming there exists a normal subgroup $H$ of order $8$ in $S_4$, As $H\unlhd G$, $HK\leq G$ for any subgroup $K\leq G$ and $|HK|=\frac{|H||K|}{|H\cap K|}$

what i am trying to do is try getting an element $ x $ of order $2$ which is not in $H$ and set $K=\{1,x\}$ then $HK$ would be a group of order $16$ which is a contradiction as $S_4$ can not have a subgroup of order $16$.

as there are six $2-cycles$ and three products of disjoint $2$ cycles but $|H|=8$, there does exists an element of order $2$ which is not in $H$ and thus we are done.

I am sure this would be the nicest way or the stupidest way one can ever do :P

I would be thankful if someone can help me to see if anything is wrong in my approach.

I would be thankful if someone can give me a hint for an alternate approach.

Thank You :)

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    $\begingroup$ This is in fact quite a nice approach in my opinion. $\endgroup$
    – user169852
    Oct 1, 2017 at 2:32

6 Answers 6

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You approach is OK, but here is a simpler one. Since $S_4/H \cong C_3$ is abelian, it follows that $[S_4,S_4]=A_4 \subseteq H$. Can you see this leads to a contradiction?

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  • $\begingroup$ +1. I do not understand how much more time will it take for me to write simple solutions than writing a one page solution..... :( I understand the contradiction, $|A_4|=12$ and $|H|=8$.. $\endgroup$
    – user87543
    Sep 5, 2013 at 14:02
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    $\begingroup$ Praphulla, do not worry! The beauty of mathematics is that different proofs can lead to the same theorem. This is how you learn mathenmatics! It happened many times that famous mathematicians went back to their/an original proof and simplified it substantially! This phenomenon warrants maybe a separate entry in this StackExchange ... $\endgroup$ Sep 5, 2013 at 14:22
  • $\begingroup$ Thanks for your concern Sir, :) $\endgroup$
    – user87543
    Sep 6, 2013 at 5:44
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    $\begingroup$ @nickyhekster what does $[S_4,S_4]$ stand for? $\endgroup$
    – GuPe
    Apr 13, 2017 at 15:44
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    $\begingroup$ The commutator subgroup, so (isomorphic to) the alternating group $A_4$. In general $[G,G]=\langle x^{-1}y^{-1}xy : x,y \in G \rangle $. $\endgroup$ Apr 13, 2017 at 15:50
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Yet another approach: Normal subgroups are comprised of entire conjugacy classes, one of which must be the identity's conjugacy class, but the class equation for $S_4$ is $$24=1+3+6+6+8,$$ and there is no way to obtain $8$ as a sum of terms from the right hand side including $1$.

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    $\begingroup$ this is useful after knowing "class equation". but i am solving problems on cosets and lagrange theorem from the book "ABSTRACT ALGEBRA-DUMMIT FOOTE". So, I can not use this approach. But i liked it. :) $\endgroup$
    – user87543
    Sep 5, 2013 at 14:12
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If you have the Sylow theorems available, then you can argue as follows. The dihedral group $D_4$ (symmetries of a square) has order $8$ and can be viewed as a subgroup of $S_4$ by regarding a symmetry of the square as a permutation of the four vertices. The permutation $p$ that interchanges two adjacent vertices of the square while fixing the other two vertices is in $S_4$ but not in $D_4$. Since it has order $2$, this $p$ must be in a $2$-Sylow subgroup. So you have at least two distinct $2$-Sylow subgroups of $S_4$, namely $D_4$ and one containing $p$. But all the $2$-Sylow subgroups are conjugate to each other, so, as soon as there's more than one of them, none of them can be normal.

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  • $\begingroup$ hayyo!! it is becoming more and more complicated :O $\endgroup$
    – user87543
    Sep 6, 2013 at 5:43
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Another approach : Two elements in $S_n$ are conjugate to each other iff they have the same cycle decomposition. Hence, all the transpositions must be in $H$ if one of them is (because $H$ is normal).

This leaves one non-identity element, which must have order 2 (since all the transpositions are their own inverses). That element must be a product of transpositions. Such a thing would be conjugate to something else, which is not in $H$.

If you assume that $H$ does not contain any transpositions, it still contains an element of order 2 ...

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  • $\begingroup$ at first sight i think i understood, but I am not sure now. Assuming $H$ has a transposition we concluded that all transpositions must be in $H$. This leaves one non identity element. I see that it has to be product of transpositions, and its conjugate has to be in $H$ which is not possible. So, H is not normal. right? $\endgroup$
    – user87543
    Sep 5, 2013 at 14:16
  • $\begingroup$ Yes, exactly. Same goes if you assume a product of two disjoint transpositions is in $H$ $\endgroup$ Sep 5, 2013 at 16:02
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    $\begingroup$ Could you please elaborate the case where $H$ does not contain transpositions. I see that it should still have an element of order 2, so an element of the type product of 2 transpositions. There are 3 of those. But then where is the contradiction? Thanks in advance $\endgroup$
    – John11
    Nov 2, 2018 at 13:35
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    $\begingroup$ @John11 yes there are three of those product of 2 transpositions...now there are remaining 4 nonidentity elements...what could they be.....they could be transpositions but then there are 6 of them so not possible....they could be 4 cycles but there are 6 of them again not possible.....they could be 3 cycles but there are 8 of them so not possible...so you see there is a problem in having a normal subgroup of size 8....what i am saying is look at the possible cycle decompositions $\endgroup$
    – RagingBull
    Dec 10, 2020 at 5:30
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I know this question is old, but there is an approach that uses one of the other problems in the same section as this one in "Abstract Algebra" by Dummit and Foote. However, it appears a little later . I thought I would just add it.

If $N$ is a normal subgroup of a finite group $G$ and $GCD(|N|, |G : N|) = 1$, then $N$ is the unique subgroup of $G$ of order $|N|$.

Using this result, assume $S_4$ has a normal subgroup, $N$, of order $8$. $GCD(8, 24/8) = 1$, thus it must be the only subgroup of order $8$. $S_4$ also has a subgroup, $K$, isomorphic to $D_8$ (dihedral group of order 8/symmetries of a square), by associating $1,2,3,4$ with vertices of square. Thus $K=N$ by the result above.

However we can show $K$ is in fact not normal. Let's say our labeling of the vertices is such that $(1,2,3,4) \in K, (1,2, 3) \notin K$ , then $(2,1,3)(1,2,3,4)(1,2,3) = (1,2,4,3) \notin K$. Thus a contradiction.

Or simply note that $S_4$ has more than one subgroup of order $8$ leading to a contradiction as well.

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Suppose $N$ is a normal subgroup of $S_4$ of order $8$. The $2$-cycles are: $(12), (13), (14), (23), (24)$ and $(34)$. So we have $6$ elements of order $2$.

Claim: There exists an element of order $2$ that is not in $N$.

Proof: If $N$ contains all the $2$-cycles of $S$, then $1, (12), (13), (14), (23), (24)$ and $(34)$ are all in $N$. Since $|N|=8$, then $N$ contains an element of order $3$ or $4$. So $N$ must also contain the inverse of that element (and the elements of order $3$ or $4$ are not self-inverse) so that $N$ will contain at least $9$ elements. This contradicts the order of $N$. Therefore there exists a $2$-cycle $(ab)\notin N$.

Now take $H=\{1,(ab)\}$. $H$ is a subgroup of $G$. Since $N$ is normal in $G$, $HN$ is a subgroup of $G$. Clearly, $H\cap N=1$ since $(ab)\notin N$. It follows that $|HN|=\frac{|H||N|}{|H\cap N|}=|H||N|=16$. But $16$ is not a divisor of $24$, contradiction Lagrange's Theorem. Therefore $S_4$ does not have a normal subgroup of order $8$.

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  • $\begingroup$ "Since $|N| = 8$, then $N$ contains an element of order 3 or 4". Why is this true? Why can't the remaining element of $N$ be a product of two transpositions in $N$, say (1 2)(3 4), which is an element of order 2? $\endgroup$
    – Junglemath
    Jun 13, 2020 at 16:36
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    $\begingroup$ As @Junglemath stated, I believe the proof of your claim is incorrect, because you forgot to consider the other elements of order 2. However, as there are 9 elements of order 2 in $S_4$ (namely the $6$ you mentioned + other $3$ elements like $(1 2)(3 4)$), you can say that there indeed is an element of order 2 not in N and then your proof would be correct (replacing $(ab)$ by $\sigma$ such that $\sigma^2 = e$). I really liked it because I'm currently reading David S. Dummit's Abstract Algebra book and it only uses results shown up until the point of the book in which this exercise is. $\endgroup$ Sep 21, 2023 at 17:44

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