# S4 subgroup of order 12

• Dec 3rd 2013, 06:36 PM
kellsbells92
S4 subgroup of order 12
Show that S4 (the symmetric group of degree 4) has a unique subgroup of order 12.

I know that A4 is that subgroup but I'm not really sure how to show that it is the unique subgroup. Help?
• Dec 4th 2013, 09:55 AM
johng
Re: S4 subgroup of order 12
Hi,
I'm tired of trying to write latex in this editor. So the hints are in an attachment. If you have problems, post them.

Attachment 29855
• Dec 4th 2013, 11:20 AM
kellsbells92
Re: S4 subgroup of order 12
thank you!
• Dec 16th 2013, 02:17 PM
Deveno
Re: S4 subgroup of order 12
We can approach this another way:

Consider the homomorphism $\text{sgn}:S_4 \to \{-1,1\}$.

If $H$ is ANY subgroup of $S_4$, then $H \cap A_4$ is also a subgroup of $S_4$. Restricting the homomorphism sgn to this subgroup yields a homomorphism from $H \cap A_4$ to {-1,1}.

Since there are only two possibilities for the image of this homomorphism, we have either:

$\text{sgn}(H \cap A_4) = \{1\}$, which implies that $H$ is a subgroup of $A_4$ (why?), or:

$\text{sgn}(H \cap A_4) = \{-1,1\}$, which implies that $H \cap A_4$ is of index 2 in $H$.

Now if $|H| = 12$, the first possibility leads to $H = A_4$. So if $H \neq A_4$, we must have that $|H \cap A_4| = |H|/[H: H \cap A_4] = 12/2 = 6$.

Now $H \cap A_4$ is thus a subgroup of $A_4$ of order 6. There are two possibilities:

a) This subgroup is cyclic, but $A_4$ has no elements of order 6, which leaves us with:

b) $H \cap A_4 \cong S_3$.

However, if $A_4$ has a subgroup isomorphic to $S_3$, this subgroup would contain 3 elements of order 2. However, the (only) 3 elements of order 2 in $A_4$, namely:

(1 2)(3 4), (1 3)(2 4), (1 4)(2 3) generate a subgroup of order 4, and $S_3$ has no such subgroup (it cannot, for 4 does not divide 6).