Abstract Algebra Quotient Group and Isomorphism Proof Help

Question

If $G$ is an abelian group, $S = \{ y \in G \; : \; y = x^2\; \exists x \in G\}$, and $T = \{ a \in G \; :\; a^2 = e\}$, then $G/T$ is isomorphic to $S$.

Proof:

Let $G$ be an abelian group, $S = \{ y \in G \; : \; y = x^2 \; \exists x \in G\}$, and $T = \{ a \in G \; : \; a^2 = e\}$.
Now $e^2 = ee = e$. So $e \in T$ and $T$ is nonempty.
Let $x,y \in T$ so $(xy^{-1})^2$ = $xy^{-1} xy^{-1} = xxy^{-1}y^{-1} = x^2 y^{-2} = x^2 (y^2)^{-1} = ee = e$.
Thus, $xy^{-1} \in T$ so $T \leq G$ by the Subgroup Test.

Let $t \in T$ and $g \in G$.
Then $(gtg^{-1})^2 = gtg^{-1} gtg^{-1} = gttg^{-1} = gt^2g^{-1} = geg^{-1} = gg^{-1} = e$.
Thus $gtg^{-1} \in T$ and $T ⊴ G$. Hence, $G/T$ is a group.

Do I have to prove $S$ is a group? Then to prove they are isomorphic to each other, I have to show a 1-1 and onto homomorphism, but I am not sure how to do that.

Answer 1

Yes, in principle you should show $S$ is a group, but this shouldn't be too difficult. The meat of the problem is exhibiting the existence of the isomorphism.

The easiest way to do this is to use the first isomorphism theorem:

Let $G$ and $G'$ be groups and let $\varphi:G\to G'$ be an onto group homomorphism with kernel $K$. Then $G/K$ is isomorphic to $G'$.

(That the kernel of a homomorphism is a normal subgroup is something you should be familiar with, or at least it's easy to prove.)

So a good strategy for this proof is to think of an onto homomorphism from $G$ to $S$, and show that its kernel is $T$. In fact, from the definition of $S$ there is a quite natural onto homomorphism to consider. (I don't want to give it away, it would be too easy then!)

Answer 2

Your notation of $S$ can be made better as $S =\{x^2: x\in G\}$

Hint :

Consider $\eta : G \rightarrow S =\{x^2: x\in G\}$ defined as $\eta(x)=x^2$

• Show that $\eta$ is a homomorphism.
• Show that $\eta$ is surjective.
• What is kernel of $\eta$.
• You remember some isomorphism theorem which is relevant to homomorphism which is surjective??

Answer 3

Proving that $S\le G$ would be a good first step. It only makes sense to talk about homomorphisms between the same types of structures. Then, the next step is to prove that $G/T \cong S$. You could, as you suggested, give an explicit isomorphism. An easier way, would be to apply the first isomorphism theorem. It will be easier to use the theorem, but, if you have not learned it yet, then you should try to define the isomorphism explicitly