Let $G$ be a finite group of order $n \geq 2$.
I want to prove that there always exists an injective homomorphism $\varphi:G \to GL_n(\mathbb R)$. Can you help?
[Math] Homomorphisms into the General Linear Group
abstract-algebrafinite-groupsgroup-theorymatrices
Related Solutions
Your claim is false. Take $G$ to be arbitrary and take $H$ to be nontrivial. let $\psi\colon G\to H$ be the trivial map, $\phi(g)=e_H$ for all $g\in G$. Then $\mathrm{Im}(\phi)=\{e\}\triangleleft H$, so the quotient exists. But the map is very nonsurjective.
More generally, if $H$ is any nontrivial group and $N\triangleleft H$ is any proper normal subgroup (they always exist, since you can take $N=\{e\}$) then the embedding $i\colon N\hookrightarrow H$ is non-surjective, but $H/\mathrm{Im}(i) = H/N$ exists.
The real reason your argument fails is that while it is true that your composition has $G$ as the kernel, the isomorphism theorem only guarantees that the image of $\psi\circ\phi$ is isomorphic to $G/\mathrm{ker}(\psi\circ\phi)$. And by construction, that image is trivial, so there is no surprise that $\mathrm{ker}(\psi\circ\phi) = G$. You have no warrant for claiming that the image is all of $G'$. You can require (but did not do so) that $G'$ be equal to the image of $\psi$, but $\mathrm{Im}(\psi\circ\phi)\subseteq \mathrm{Im}(\psi)$, and equality need not hold. We know the equality always holds when $\phi$ is surjective... but that is what you are trying to prove, so you cannot assume it holds.
For abelian group, and more generally for modules, there is the dual concept to the kernel called the cokernel; given $f\colon M\to N$, the cokernel of $f$ is $\mathrm{coker}(f)=N/\mathrm{Im}(f)$. It is in fact the case that $f$ is surjective if and only if the cokernel is trivial, just like $f$ is injective if and only if the kernel is trivial. This does not work for arbitrary groups, though, because the image need not be normal. If you quotient out by the normal closure of the image you get the equivalent construction but it no longer "detects" surjectivity (because surjectivity is not a categorical notion).
For finite groups, in the spirit of of Ravi's comments, you can prove the following result in an elementary way (without using the theory of injective modules or the structure theorem for a finite abelian groups):
Thm. Let $G$ be a finite abelian group, and let $H$ be a subgroup. Then any morphism $H\to\mathbb{C}^*$ extends to a morphism $G\to\mathbb{C}^*$.
The proof is by induction on $m=[G:H]$. If $m=1$, it's OK. If $>1$, pick $g\in G\setminus H$, and let $n$ the order of $gH$ in $G/H$. In other words, $n$ is the smallest positive integer such that $g^n\in H$. Let $\chi:H\to\mathbb{C}^*$ a morphism, and set $t=\chi(g^n)$. We can write $t=w^n$ for some $w\in\mathbb{C}^*$. Let $H'=\langle H,g\rangle.$ Now every element $h'$ of $H'$ can be written $h'=hg^k,h\in H,k\in\mathbb{Z}$. Set $\chi'(h')=\chi(h) w^k$. One can show that this does not depend on the decomposition of $h'$ and that $\chi'$ is a morphism. Now $g\in H'\setminus H$, so $[G:H']<[G:H]$ and we may apply induction.
Now to solve your question, since $x$ has order $4$, we have a morphism $H=\langle x\rangle \to\mathbb{C}^*$ sending $x$ to $i$, and we can apply the theorem.
Best Answer
By Cayley's theorem, we have an injective homomorphism, $$G\hookrightarrow S_n.$$ We have a natural action of $S_n$ on $\mathbb{R}^n$ by permuting the indices via: $$S_n \times \mathbb{R}^n \to \mathbb{R}^n, \ (\sigma,(x_1,...,x_n))\mapsto (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$ This action induces a homomorphism $$\psi :S_n \to \mbox{Aut}(\mathbb{R}^n)=GL_n(\mathbb{R}), \ \psi(\sigma)((x_1,...,x_n))= (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$
You can easily check that the map thus defined is indeed a homomorphism. The claim follows.
I hope this helps!