Recent content by ttm7nana

Okay, that makes sense. Thank you again for your help!

Okay, thank you very much! If I may ask, what was your thought process in first showing ab\notinH iff ba\notinH? I would have never thought of that step on my own.

Hi, thanks for your help and the welcome! I think I've got it. Can you check my work? Let ab\notinH. From this we get Ha ≠Hb^(-1). Using Ha≠Hb then aH≠bH, aH ≠ b^(-1)H. This implies that ba\notinH. And the other direction is the same procedure. Therefore ab\notinH iff ba\notinH Now assume...

Hello, I am having trouble with the following problem. Suppose that H is a subgroup of G such that whenever Ha≠Hb then aH≠bH. Prove that gHg^(-1) is a subset of H. I have tried to manipulate the following equation for some ideas H = Hgg^(-1) = gg^(-1)H but I don't know how to go...