When we want to prove that a subgroup is normal, we usually show that the left and right cosets are equal, right? But what if a group has so many elements, is not abelian, and the subgroup does not have index 2…how can we show that such a subgroup is normal?
Thanks in advance
Best Answer
There are many ways of showing a subgroup is normal. One would be to show that the subgroup is the kernel of some homomorphism. There are some general theorems that guarantee a subgroup of a finite group to be normal, by some combinatorial argument. For instance, if $H$ has index $p$ and $p$ is the smallest prime that divides $|G|$, then $H$ must be normal. (This extends the easy result that a subgroup of index $2$ is normal).
So, when given a particular subgroup and asked if it is normal you can try any of a rather limited number of tricks, or resort to (any of the equivalent) definitions of normality.