Prove group and opposite group are isomorphic.

Question

Let $\langle G, ∗\rangle$ be a group, and suppose $∗'$ is a binary operation on $G$ defined by $a∗'b=b∗a$ for all $a,b \in G$. I also proved in the previous problem that $\langle G, ∗'\rangle$ was a group. What I'm thinking is:

Proof: Let $\langle G, ∗\rangle$ be a group, and suppose $∗'$ is a binary operation on $G$ defined by $a∗'b=b∗a$ for all $a,b \in G$. From the previous example we know that $\langle G, ∗'\rangle$ was a group. We know that $\langle G, ∗'\rangle$ contains all elements of that $\langle G, ∗\rangle$ since the only difference is the operation. Therefore, the groups have the same order, and thus are 1-to-1. Given that $*'$ is defined as $a∗'b=b∗a$ for all $a,b \in G$, this is essentially the commutative property to $*$. So, $a∗'b=b∗a$ and $b∗'a=a∗b$ thus making it an onto mapping. Therefore, they are bijective. Hence, $\langle G, ∗\rangle$ and $\langle G, ∗'\rangle$ are isomorphic as groups.

Is my reasoning sound?

Answer 1

Proof: Let $\langle G, ∗\rangle$ be a group, and suppose $∗'$ is a binary operation on $G$ defined by $a∗'b=b∗a$ for all $a,b \in G$. From the previous example we know that $\langle G, ∗'\rangle$ was a group. We know that $\langle G, ∗'\rangle$ contains all elements of that $\langle G, ∗\rangle$ since the only difference is the operation. Therefore, the groups have the same order, and thus are 1-to-1.

When you say 1-to-1, you need first give a map, $\phi:G\to G'$, and show $\phi$ is 1-to-1. How is your $\phi$ defined here?

Given that $*'$ is defined as $a∗'b=b∗a$ for all $a,b \in G$, this is essentially the commutative property to $*$. So, $a∗'b=b∗a$ and $b∗'a=a∗b$ thus making it an onto mapping. Therefore, they are bijective. Hence, $\langle G, ∗\rangle$ and $\langle G, ∗'\rangle$ are isomorphic as groups.

The general step to prove $G$ and $G'$ are isomorphic is:

1. give a map $\phi: G\to G'$
2. show $\phi$ is one to one
3. show $\phi$ is onto
4. show $\phi$ is a homomorphism, namely, $\phi(a*b)=\phi(a)*'\phi(b), ~~\forall a,b\in G$

Answer 2

The map $x \mapsto x^{-1}$ is an isomorphism from $(G,*)$ to $(G,*')$.

Answer 3

The identity mapping, although bijective, doesn't respect the group structure. So, you need to come up with a different bijection which is also a homomorphism of groups.

Hint : The new operation, although drastically different from the old one, does have the same identity element and the same inverse for each element.

How about the map $x \mapsto x^{-1}$ ?

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As a side note, if you are aware of a bit of category theory (ignore, if not), then the new group is just the dual/opposite of $G$, considered as a single object groupoid. So what this says, is that a single object groupoid is isomorphic (as a category) to it's dual.