Let $A$ be a group and $a \in A$. Also, $H$ is a subgroup of $A$ as well as $K= \{ aha^{-1}\mid h\in H \}$. How to show that $K$ is isomorphic to $H$?

Question

Let $A$ be a group and $a \in A$. Also, $H$ is a subgroup of $A$ as well as $K= \{ aha^{-1}\mid h \in H \}$. How to show that $K$ is isomorphic to $H$?

So I know that I need to define a function $f$ that is both bijective and satisfies the operation-preserving property. That is, $f : H \to K$ given by $f(h)=aha^{-1}$.

It is easy show that $f$ is injective and has the operation-preserving property, but I couldn't show that it is surjective. My way of showing that it is surjective is to take $y \in K$, then we could write $y$ as $y=aha^{-1}$ for some $h \in H$. Then $f(h)= y$.

But I am not convinced that this is logically correct. Is there a better way on showing the surjectivity? I tried interchanging the domain and range, that is, $f:K \to H$. But that would result to a more complicated scenario.

The map is called conjugation and it is well known to be an isomorphism. — Shaun, Mar 26 '23 at 16:28
You may show that $g:K\to H$, $g(y) = a^{-1} ya$ is the inverse of $f$. — Arctic Char, Mar 26 '23 at 16:29

score -3 · Accepted Answer · answered Mar 26 '23 at 16:33

-3

The way you are doing is correct. Secondly if $A$ is finite group then clearly cardinality of both $K$ and $H$ is same. And this implies $f$ is surjective.

answered Mar 26 '23 at 16:33

Sunil Kumar

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Please do not answer such FAQs; instead, vote to close them as duplicates. – Shaun Mar 26 '23 at 16:38

Let $A$ be a group and $a \in A$. Also, $H$ is a subgroup of $A$ as well as $K= \{ aha^{-1}\mid h\in H \}$. How to show that $K$ is isomorphic to $H$?

1 Answers1