Question

Suppose that G is a simple group and $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$is a surjective homomorphism of groups. Prove that either f is an isomorphism or$\mathit{H}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$ .

Short answer

It is proved that, either f is an isomorphism or$H=⟨e⟩$ .

Step-by-step solution

Step 1:Important Theorems

Theorem 8.11

The following conditions on a subgroup$\mathit{N}$ of a group$\mathit{G}$ are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every$a\in \text{\hspace{0.17em}\hspace{0.17em}}G$ , Where$a^{-1}Na\subseteq N$$\left\{a^{-1}Na|n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every$a\in \text{\hspace{0.17em}\hspace{0.17em}}G$, Where$aN{a}^{-1}\subseteq N,$$\left\{aN{a}^{-1}|n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every$a\in \text{\hspace{0.17em}\hspace{0.17em}}G$.
5. $aN{a}^{-1}\subseteq N,$for every$a\in \text{\hspace{0.17em}\hspace{0.17em}}G$ .

Theorem 8.18

If N is a normal subgroup of a group G, then map$\pi :G\to G/N$ given by$\pi \left(a\right)=Na$ is a surjective homomorphism with kernel N.

Proving that either f is an isomorphism orH=⟨e⟩

As it is given that, G is a simple group and .$f:G\to H$

According toTheorem 8.18, if is a normal subgroup G, then G and H

are homomorphic.

Therefore, first we must proof that$\ker f$ is a normal group.

Taking$a\in kef$ and $g\in G$, find $f\left(ga{g}^{-1}\right)$as:

$\begin{array}{c}f\left(ga{g}^{-1}\right)=f\left(g\right)f\left(a\right)f{\left(g\right)}^{-1}\\ =f\left(g\right).e_H.f{\left(g\right)}^{-1}\\ =e_H\end{array}$

Thus,$ga{g}^{-1}\in \ker f$, which implies that $ga{g}^{-1}\subseteq \ker f$.

So, from above result and byTheorem 8.11, we can conclude that$\ker f$ is a normal subgroup of G.

Since G is a simple group, $\ker f$is either $\left\{e_G\right\}$or G.

If$\left\{e_G\right\}$ , then f is an injective homomorphism.

And, if $\ker f=G$, then $H=⟨e_g⟩$and fshould be surjective.

Since it is already given that f is surjective, either f is an isomorphism or$H=⟨e⟩$ .

Hence, we can conclude that either f is an isomorphism or$H=⟨e⟩$ .