Let G be a group H be a subgroup of G which has exactly two distinct cosets.Let C={H'⊂G :H'=gHg^-1 for some g∈G} How many elements Of C have?

Question

Let $G$ be a group and let $H$ be a subgroup of $G$ which has exactly two distinct cosets. Let $$C=\left\{H'⊂G :H'=gHg^{-1} \text{ for some } g∈G\right\}$$

How many elements does $C$ have?

You've been here 16 months and posted 29 previous questions. You should know how to post a much higher quality question. — , Sep 05 '17 at 15:21

score 1 · Answer 1 · answered Sep 05 '17 at 15:22

1

$[G:H]=2$ so $H$ is normal in $G$ (see 1, 2).

Therefore, it has exactly one conjugate, namely itself.

answered Sep 05 '17 at 15:22

Kenny Lau

25,049

Let G be a group H be a subgroup of G which has exactly two distinct cosets.Let C={H'⊂G :H'=gHg^-1 for some g∈G} How many elements Of C have?

1 Answers1