# cosets

• Oct 25th 2009, 10:44 PM
sfspitfire23
cosets
hi all, just an interesting coset problem I'm having trouble with -

Find the right and left cosets in G of the subgroups H and K of G.

$G=A_4$; $H=\{e,(1 2)(3 4),(1 3)(2 4), (1 4)(2 3)\}$; $K=<(1 2 3)>$

What do you guys think?
• Oct 26th 2009, 02:56 AM
Swlabr
Quote:

Originally Posted by sfspitfire23
hi all, just an interesting coset problem I'm having trouble with -

Find the right and left cosets in G of the subgroups H and K of G.

$G=A_4$; $H=\{e,(1 2)(3 4),(1 3)(2 4), (1 4)(2 3)\}$; $K=<(1 2 3)>$

What do you guys think?

Firstly, you know precisely how many right/left cosets there are for each subgroup, this is just $|G|/|A|$ where $A$ is your subgroup.

Now, for left cosets we know $xA=yA \Leftrightarrow x^{-1}y$, and we know that cosets all have equal size (they have size $|A|$). Thus, you need to find however many sets of 4 elements not in $H$ such that $x^{-1}y \in H$ for all 4 elements.

These will be your left cosets. For your right cosets, you apply a similar formula ( $xy^{-1}$).

Remember that left and right cosets are equal if and only if your subgroup is $Normal$. I'm afraid I can't tell you off hand whether $K$ is normal, although I am pretty positive it is. $H$ is definitely normal. So, hopefully, in each case your left and right cosets will be equal.
• Oct 26th 2009, 09:30 AM
sfspitfire23
Would e*k={e, (123),(132)}?

I think this would be correct...but why is the (132) in there?

thanks
• Oct 27th 2009, 01:56 AM
Swlabr
Quote:

Originally Posted by sfspitfire23
Would e*k={e, (123),(132)}?

I think this would be correct...but why is the (132) in there?

thanks

This is correct - the coset $eH$ is just $H$, as when you multiply every element in the subgroup $H$ by the identity you get the identity. Similarly, $hH=H$ for all $h \in H$. $(132)$ is in the coset because $K=\{e, (123), (132)\}=<(123)>$. The $$ brackets mean "the subgroup generated by the set $X$". Have you come across this notation before?
• Oct 27th 2009, 04:51 PM
sfspitfire23
ah indeed! I see now!