By H6, I assume you mean the symmetric group on 6 elements. I've seen that denoted or , but never .
With that assumption, your reasoning and answer are correct.
Hi anyone knows how to solve this?
find the number of distinct left cosets of H ={e, (16), (23), (16)(23)} in H_{6}?
My workings as below,
Since H_{6}, 6!= 1x2x3x4x5x6=720
Since H has 4 elements, i take 720/4=180 (Answer)
Please correct me if i am wrong.
Thanks.