Let be a Subgroup of the Group , of [index_of_subgroup index] . Then is a Normal subgroup of .
Proof
We must show that is closed under conjugation by elements of .
Since has index in , there are two left cosets: and for some specific . There are also two right cosets: and .
Now, since , it must be the case that ; so without loss of generality, .
Hence and so .
It remains to show that is closed under conjugation by every element of . But every element of is either in , or in ; so it is either or , for some .
- is equal to since and .
- .
This completes the proof.