Show that $H$ is a group, and that it is isomorphic to $G$

Question

Let $G$ be any group. Let $H$ consist of the same set of elements as $G$, but with a new operation given by $a ∗ b = ba$, for all $a$ and $b$. Show that $H$ is a group, and that it is isomorphic to $G$.

I am having trouble proving what is being asked of me. First, I have no information about how the operation is in $G$. I suppose I could define the operation in $G$ as is usually done for group definition, this is if $a, b \in G$, then its product is $ab$. Based on that, $H$ with the new operation would be a group as well, since they have the same elements. The truth seems like nonsense to me but I can't think of anything else.

Also I have tried to establish an isomorphism and let the rest follow from this fact, but so far it has been unsuccessful. Any hint would be helpful.

You can show this directly from the axioms. For example can you see why $H$ is closed under $$ using the given definition for $$ in terms of the group operation for $G$? — Noah Solomon, Feb 03 '21 at 18:47
It isn't the same group, because the operation is different. A group comprises a set of elements and a binary operation. — saulspatz, Feb 03 '21 at 18:49
As a hint, try mapping elements $a$ in the group to $a^{-1}$. — Derek Holt, Feb 03 '21 at 18:56
Does this answer your question? Prove that $G^{op} \cong G$ when $G^{op}$ is the opposite group of G — , Feb 03 '21 at 19:42

score 2 · Accepted Answer · answered Feb 03 '21 at 18:59

You should be able to check the axioms of group for $H$. For instance if $e$ is the neutral element of $G$ then so is for $H$ since $e * g = ge=g$ and if $g^{-1}$ is the inverse of $g$ in $G$ then so is it in $H$ since $g * g^{-1}= g^{-1} g =e$.

For the isomorphism define $f: G \to H$, $f(g):= g^{-1}$. It is a group homomorphism since $$f(ab)=(ab)^{-1}= b^{-1}a^{-1} = f(b)f(a)= f(a)*f(b).$$ Its inverse $f^{-1}: H \to G$ is defined exactly in the same way, $f^{-1}(g):=g^{-1}$, and it is indeed the inverse of $f$ since $(g^{-1})^{-1}=g$.

Show that $H$ is a group, and that it is isomorphic to $G$

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