Recent questions in Abstract algebra

Recent questions in Abstract algebra
stop2dance3l 2022-01-15 Answered

Let G be a group and H a subgroup of G. Show that \(\displaystyle{a}{H}={H}{a}\) if and only if \(\displaystyle{a}{h}{a}^{{-{1}}}\in{H}\) for every \(\displaystyle{h}\in{H}\).
Suppose that \(\displaystyle{a}{H}={H}{a}\). Then for \(\displaystyle{h}'\in{a}{H}\) we have that \(\displaystyle{h}'={a}{h}\) for some \(\displaystyle{h}\in{H}\). Right multiplying by \(\displaystyle{a}^{{-{1}}}\) we have that \(\displaystyle{h}'{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}\). As \(\displaystyle{a}{H}={H}{a}\) we have that \(\displaystyle{h}'={a}{h}={h}{a}\) so \(\displaystyle{h}'{a}^{{-{1}}}=\underbrace{{{h}{a}{a}^{{-{1}}}}}={a}{h}{a}^{{-{1}}}\Rightarrow{a}{h}{a}^{{-{1}}}\in{H}\).
Conversely suppose that \(\displaystyle{a}{h}{a}^{{-{1}}}\in{H}\). The claim that \(\displaystyle{a}{H}={H}{a}\) so we will do both inclusions. Let \(\displaystyle{h}'\in{a}{H}\). Then \(\displaystyle{h}'={a}{h}\) for some \(\displaystyle{h}\in{H}\). Now right multiplying by \(\displaystyle{a}^{{-{1}}}\) we ahve that \(\displaystyle{h}'{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}\in{H}\). So multiplying again from right I have that \(\displaystyle{h}'{a}^{{-{1}}}{a}={a}{h}{a}^{{-{1}}}{a}\Rightarrow{h}'={a}{h}\in{H}{a}\).
For the other direction let \(\displaystyle{h}{''}\in{H}{a}\). Then \(\displaystyle{h}{''}={h}{a}\) for some \(\displaystyle{h}\in{H}\). Right multiplying by \(\displaystyle{a}^{{-{1}}}\) then left multiplying by a and again right multiplying by \(\displaystyle{a}^{{-{1}}}\) I get that \(\displaystyle{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}\in{H}\). Now left multiplying by a I have that \(\displaystyle{a}{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={a}{a}{h}{a}^{{-{1}}}\in{a}{H}\).
Is it necessarily true that \(\displaystyle{a}{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={h}{''}\) and that \(\displaystyle{a}{a}{h}{a}^{{-{1}}}={a}{h}\)? This would seem to conclude the result, but I'm not sure if it's allowed.
Also is should I do these kind of ''multiplying'' here? It seems that I am going in circles multiplying everywhere.

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