Show that S4 has a unique subgroup of order 12.

**nhk** · April 20th 2010, 10:22 PM

I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?

**NonCommAlg** · April 21st 2010, 12:47 AM

Originally Posted by nhk

I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?

first show that $A_4$ has no subgroup of order $6$ . then, since $|HA_4|=\frac{144}{|H \cap A_4|} \mid 24$ and $|H \cap A_4| < 12,$ we must have $|H \cap A_4|=6.$ contradiction!

**Swlabr** · April 21st 2010, 01:00 AM

Originally Posted by nhk

I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?

I don't know how much you done have on group theory already, but the subgroup you are dealing with has index 2, and is denoted $A_4$ . It is called the alternating group of degree 4. the question asks you to assume the subgroup is normal because it is; every subgroup of index 2 in a group is normal (essentially, because there can only exist two cosets, and so left and right cosets must be equal).

Now, permutations in $S_n$ are special. A permutation of a specific cycle type* is conjugate to every other permutation of that cycle type. (g is conjugate to h if there exists some element a such that $g = a^{-1}ha$ ). For example, $(12)(123)(12) = (132)$ . This should hopefully be in your notes. If it is not, then say and someone will have to come up with a different method to solve this probelm!

A normal subgroup is a group which is closed under conjugation. So in $S_n$ , if there exists a permutation of a cycle type then all permutations of this cycle type are in the normal subgroup.

Let us now count the number of each cycle types,

There is 1 cycle of type $()$ in $S_4$ .

There are 4.3/2 = 6 cycles of type $(12)$ in $S_4$ .

There are 4.3.2/3 = 8 cycles of type $(123)$ in $S_4$ .

There are 4.3.2.1/4 = 6 cycles of type $(1234)$ in $S_4$ .

There are 4.3.2.1/(2.2.2) = 3 cycles of type $(12)(34)$ in $S_4$ .

(for a normal cycle you divide by the length as $(12) = (21)$ , $(123) = (231) = (312)$ etc. For the last part you divide by 2.2.2 as two of the twos are because $(12)=(21)$ and the third is because $(12)(34) = (34)(12)$ . Alternatively, just sum up all the other cycle types and subtract from 24, 24-(6+8+6+1) = 3.)

Note that these all sum up to 24. Now, we wish to sum together the numbers 1, 6, 8, 6 and 3 to get 12 in two different way. However, there is only one way of doing this. This means there can only be one subgroup of order 12 in $S_4$ .

*Note that by cycle type I mean $(123)(45)$ , $(123)$ , $(12)(34$ are all different while $(123)$ and $(132)$ have the same cycle type.

**Swlabr** · April 21st 2010, 01:02 AM

Originally Posted by NonCommAlg

first show that $A_4$ has no subgroup of order $6$ . then, since $|HA_4|=\frac{144}{|H \cap A_4|} \mid 24$ and $|H \cap A_4| < 12,$ we must have $|H \cap A_4|=6.$ contradiction!

Do you mean normal subgroup? As there is an obvious copy of $A_3$ in there...

**NonCommAlg** · April 21st 2010, 11:21 AM

Originally Posted by Swlabr

Do you mean normal subgroup? As there is an obvious copy of $A_3$ in there...

the order of $A_3$ is $3$ not $6$ .

**Swlabr** · April 21st 2010, 12:17 PM

Originally Posted by NonCommAlg

the order of $A_3$ is $3$ not $6$ .

Good point, not sure why I read it as $S_4$ .

**nhk** · April 26th 2010, 08:15 AM

Originally Posted by NonCommAlg

first show that $A_4$ has no subgroup of order $6$ . then, since $|HA_4|=\frac{144}{|H \cap A_4|} \mid 24$ and $|H \cap A_4| < 12,$ we must have $|H \cap A_4|=6.$ contradiction!

sorry for taking so long to respond, but why is it that we can use those set of equations that you have up there, is it because both of the subgroups of order 12 are normal?
Thanks again for your help, it has been greatly appreciated.

**Swlabr** · April 26th 2010, 09:28 AM

Originally Posted by nhk

sorry for taking so long to respond, but why is it that we can use those set of equations that you have up there, is it because both of the subgroups of order 12 are normal?
Thanks again for your help, it has been greatly appreciated.

Do you mean the $|AB| = \frac{|A|\cdot|B|}{|A \cap B|}$ ? This is a well-known formula and holds if A, B and AB are all subgroups.

**nhk** · April 26th 2010, 10:33 AM

Originally Posted by Swlabr

Do you mean the $|AB| = \frac{|A|\cdot|B|}{|A \cap B|}$ ? This is a well-known formula and holds if A, B and AB are all subgroups.

THanks, i thought that formula was valid but I could'nt find it in my book

. Must have been looking in the wrong section. Thanks for your help and the fast response

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