# Thread: Show a subgroup is normal

1. ## Show a subgroup is normal

Hey everyone,

I have an exercise in which I need to show the subgroup $\displaystyle H \subset S_4$ is normal, when $\displaystyle H$ is given as:

$\displaystyle H = \lbrace e, (12)(34),(13)(24),(14)(23) \rbrace$

I've maneged to show $\displaystyle H$ is indeed a subgroup and my idea is to somehow use the conjugate of a subgroup, but can't seem to make it work. Is there a better way to tackle this problem?

2. ## Re: Show a subgroup is normal

Originally Posted by Krisly
I have an exercise in which I need to show the subgroup $\displaystyle H \subset S_4$ is normal, when $\displaystyle H$ is given as:
$\displaystyle H = \lbrace e, (12)(34),(13)(24),(14)(23) \rbrace$
I've maneged to show $\displaystyle H$ is indeed a subgroup and my idea is to somehow use the conjugate of a subgroup, but can't seem to make it work. Is there a better way to tackle this problem?
I don't know of any quick solution for this task.
There is a theorem from which you may be able to adapt proof:
A subgroup $\mathscr{S}$ of a group $\mathscr{G}$ is normal iff all of its right-cosets are left-cosets.

3. ## Re: Show a subgroup is normal

A fact which is not hard to prove: in a symmetric group, a conjugate of a cycle is another cycle of the same length. So for example $g(1\,2)(3\,4)g^{-1}=g(1\,2)g^{-1}g(3\,4)g^{-1}$, another product of two 2 cycles. Now notice in $S_4$, the only products of two 2 cycles are the non-identity elements of H.

4. ## Re: Show a subgroup is normal

Thank you very much! does that mean I can argue the following using both comments?

The group $\displaystyle H$ is normal iff the left cosets equals the right cosets hence;

$\displaystyle f H = H f, \: \forall f \in S_4$, thus $\displaystyle H$ is normal iff $\displaystyle fhf^{-1} \in H \: \forall f \in S_n \land h \in H$.

And as $\displaystyle f(ab)(cd)f^{-1} = f (ab) f^{-1} f (cd) f^{-1} = (f[a]f[b])(f[c]f[d])$

in addition to the argument all composition of disjoint two cycles of $\displaystyle S_4$ belongs to the set $\displaystyle H$, then $\displaystyle fHf^{-1} \in H$, thus $\displaystyle H$ is normal?

5. ## Re: Show a subgroup is normal

Yes. As Plato points out a subgroup H of a group G is normal iff for all g in G, gH=Hg.
Now for any g in G, gH=Hg iff $gHg^{-1}=H$ iff for any g in G and any h in H
$ghg^{-1}\in H$.