# Groups

• Feb 3rd 2010, 05:53 AM
slevvio
Groups
Hello, I was wondering if I could get some help with this question.

Let G be the subgroup of $\displaystyle S_4$ generated by (1 2),(3 4) and (1 3)(2 4). Let H be the subgroup of S_4 generated by (1 2) and (3 4). Show that H is a normal subgroup of G. Show that H has order 4 and G/H has order 2. Deduce that G has order 8.

I was wondering if there is a quick way to show that G has order 4 without doing it explicitly? I was also wondering how to show that G/H has order 4. I guess that Lagrange's Theorem proves the last part. Any help with this would be appreciated.

:)
• Feb 3rd 2010, 08:15 AM
Drexel28
Quote:

Originally Posted by slevvio
Hello, I was wondering if I could get some help with this question.

Let G be the subgroup of $\displaystyle S_4$ generated by (1 2),(3 4) and (1 3)(2 4). Let H be the subgroup of S_4 generated by (1 2) and (3 4). Show that H is a normal subgroup of G. Show that H has order 4 and G/H has order 2. Deduce that G has order 8.

I was wondering if there is a quick way to show that G has order 4 without doing it explicitly? I was also wondering how to show that G/H has order 4. I guess that Lagrange's Theorem proves the last part. Any help with this would be appreciated.

:)

Clearly we have that $\displaystyle G\leqslant S_4$ and $\displaystyle H\leqslant G$.

To see that $\displaystyle \left|H\right|=4$ we must merely note that $\displaystyle \left|(1,2)\right|=\left|(3,4)\right|=\left|(1,2)( 3,4)\right|=2$. I leave it to you to prove that $\displaystyle \left|G\right|=8$ (shouldn't be too hard). It is clear that $\displaystyle H\lhd G$ since $\displaystyle \left[G:H\right]=2$. So then clearly $\displaystyle \left|G/H\right|=\left[G:H\right]=2$
• Feb 3rd 2010, 10:12 AM
slevvio
I dont understand how you've shown that H = 4 ?
• Feb 3rd 2010, 12:27 PM
Drexel28
Quote:

Originally Posted by slevvio
I dont understand how you've shown that H = 4 ?

Clearly we have that $\displaystyle H$ must contain at least $\displaystyle \left\{e,(1,2),(3,4),(1,2)(3,4)\right\}$. But, it is clear to see that $\displaystyle \left((1,2)(3,4)\right)^n=\begin{cases} e & \mbox{if} \quad n\text{ is even} \\ (3,4) & \mbox{if} \quad n\text{ is odd}\end{cases}$ and similarly for the other two elements and so it remains to check that the multiplication of any combination of them stays within the group. To do this note that $\displaystyle \left(1,2\right)^m\left(3,4\right)^n\left((1,2)(3, 4)\right)^\ell=\left(1,2\right)^m(3,4)^n (1,2)^\ell (3,4)^\ell=(1,2)^{m+\ell}(3,4)^{n+\ell}$...so
• Feb 3rd 2010, 01:20 PM
Drexel28
P.S. $\displaystyle H\cong \mathbb{Z}_2\times\mathbb{Z}_2$
• Feb 3rd 2010, 04:29 PM
slevvio
Ah I understand now. Obviously those 4 elements are in H, but since they all have order 2 you cannot get any more members when you combine them.

Thanks for the help
• Feb 3rd 2010, 04:32 PM
Drexel28
Quote:

Originally Posted by slevvio
Ah I understand now. Obviously those 4 elements are in H, but since they all have order 2 you cannot get any more members when you combine them.

Thanks for the help

Mainly because this subgroup happened to be abelian since disjoint cycles commute.
• Feb 3rd 2010, 04:33 PM
slevvio
Is there a way to show that H is normal to G without having to work out the elements and conjugate them one by one?
• Feb 3rd 2010, 04:38 PM
slevvio
Actually what I did was let X = {(1 2) , (3 4)} and let Y = { (1 2), (3 4) , (1 3)(2 4) } and showed $\displaystyle yXy^{-1} = X$.
• Feb 3rd 2010, 04:39 PM
Drexel28
Quote:

Originally Posted by slevvio
Is there a way to show that H is normal to G without having to work out the elements and conjugate them one by one?

Yes. I told you that since it has index two it's normal.

To see this you need to know that a group is normal if every left coset is a right coset. And if a subgroup has index two this is easy to see since we have that the set of left cosets looks like $\displaystyle \left\{H,\mathcal{L}\right\}$ and the set of right cosets looks like $\displaystyle \left\{H,\mathcal{R}\right\}$ and since $\displaystyle H\cup\mathcal{L}=G=H\cup\mathcal{R}$ and $\displaystyle H\cap \mathcal{L}=\varnothing=H\cap\mathcal{R}$ it followst hat $\displaystyle \mathcal{L}=\mathcal{R}$.
• Feb 3rd 2010, 04:58 PM
slevvio
OK I will think on what you have said

However the question is asking me to show that | G / H | = 2 and use that to deduce |G| = 8. It's easy to show this the other way round but I can't quite work this out
• Feb 3rd 2010, 04:59 PM
Drexel28
Quote:

Originally Posted by slevvio
OK I will think on what you have said

However the question is asking me to show that | G / H | = 2 and use that to deduce |G| = 8. It's easy to show this the other way round but I can't quite work this out

If $\displaystyle H\lhd G$ then $\displaystyle \left|G/H\right|=\left[G:H\right]$
• Feb 3rd 2010, 05:00 PM
slevvio
ah of course. It's very late here in Scotland :)
• Feb 3rd 2010, 05:13 PM
slevvio
I understand that [G:H] = | G / H | if H is normal to G and I know that |G| = 8.

But how do I show that [G:H] = 2 only knowing that H has order 4. The question does not ask me to work out |G| but to deduce it from the fact that [G:H] = 2, but how do I show this without showing |G| = 8 first.

I am trying to show that there must only be 2 distinct cosets of H in G but Im not getting very far (Sleepy)
• Feb 3rd 2010, 05:29 PM
Drexel28
Quote:

Originally Posted by slevvio
I understand that [G:H] = | G / H | if H is normal to G and I know that |G| = 8.

But how do I show that [G:H] = 2 only knowing that H has order 4. The question does not ask me to work out |G| but to deduce it from the fact that [G:H] = 2, but how do I show this without showing |G| = 8 first.

I am trying to show that there must only be 2 distinct cosets of H in G but Im not getting very far (Sleepy)

You have that $\displaystyle H\cup H(1,3)(3,4)=G$ and they are disjoint...so they are the only distinct cosets.