Let G be a group and H a subgroup of G. Show that aH=Ha if and only if aha−1∈H for every h∈H.Suppose that aH=Ha. Then for h′∈aH we have that h′=ah for some h∈H. Right multiplying by a−1 we have that h′a−1=aha−1. As aH=Ha we have that h′=ah=ha so h′a−1=haa−1⏟=aha−1⇒aha−1∈H.Conversely suppose that aha−1∈H. The claim that aH=Ha so we will do both inclusions. Let h′∈aH. Then h′=ah for some h∈H. Now right multiplying by a−1 we ahve that h′a−1=aha−1∈H. So multiplying again from right I have that h′a−1a=aha−1a⇒h′=ah∈Ha.For the other direction let h″∈Ha. Then h″=ha for some h∈H. Right multiplying by a−1 then left multiplying by a and again right multiplying by a−1 I get that ah″a−1a−1=aha−1∈H. Now left multiplying by a I have that aah″a−1a−1=aaha−1∈aH.Is it necessarily true that aah″a−1a−1=h″ and that aaha−1=ah? 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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Let G be a group and H a subgroup of G. Show that aH=Ha if and only if aha−1∈H for every h∈H.Suppose that aH=Ha. Then for h′∈aH we have that h′=ah for some h∈H. Right multiplying by a−1 we have that h′a−1=aha−1. As aH=Ha we have that h′=ah=ha so h′a−1=haa−1⏟=aha−1⇒aha−1∈H.Conversely suppose that aha−1∈H. The claim that aH=Ha so we will do both inclusions. Let h′∈aH. Then h′=ah for some h∈H. Now right multiplying by a−1 we ahve that h′a−1=aha−1∈H. So multiplying again from right I have that h′a−1a=aha−1a⇒h′=ah∈Ha.For the other direction let h″∈Ha. Then h″=ha for some h∈H. Right multiplying by a−1 then left multiplying by a and again right multiplying by a−1 I get that ah″a−1a−1=aha−1∈H. Now left multiplying by a I have that aah″a−1a−1=aaha−1∈aH.Is it necessarily true that aah″a−1a−1=h″ and that aaha−1=ah? This would seem to conclude the result, but I&#039;m not sure if it&#039;s allowed.Also is should I do these kind of &#039;&#039;multiplying&#039;&#039; here? It seems that I am going in circles multiplying everywhere.

maul124uk · Accepted Answer

As aH=Ha we have that h′=ah=ha...Unfortunately this does not follow: we simply have an equality of the sets aH=Ha. We could fix it up like this:Suppose that aH=Ha and fix h∈H. We need to show that aha−1∈H. We at least know that ah∈aH. Since aH=Ha this means that ah∈Ha and so there exists h′∈H such that ah=h′a. Thus aha−1=h′, so in particular we can conclude aha−1∈H, as we desire.Your argument for the containment aH⊂Ha seems good to me. As explained in comments, there is a problem with the other containment, which indeed does not hold unless in the statement we are qualifying over all a in G at once (and kabenyuk gives a counterexample in this case in another answer).

Kirsten Davis · Accepted Answer

There is an error at the beginning, when you write h′=ah=ha, as this implies that a commutes with h. Further, it seems overly complicated.Here&#039;s a hint: the hypothesis aH=Ha means that for every h∈H, there exists h′∈H such that ah=h′a.Hence aha−1=h′ indeed belongs to H. The converse is obvious.

alenahelenash · Accepted Answer

The converse is not true. That is, from the condition that aha−1∈H for all h∈H it does not following that aH=Ha. Here is an example. Let G=(QQ01),H=(1Z01),a=(2001) It is not difficult to check that aha−1∈H for all h∈H, but aH=(22Z01)≠(2Z01)=Ha.

Let G be a group and H a subgroup of G. Show that aH=Ha if and only if aha^{-1} \in H fo

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