Questions about an exercise about the irreducible adjoint representation The following exercise I want to ask is

Show that if every adjoint representation $G$ is irreducible, there
is no nontrivial subspace $V \subset T_{e}G$ is invariant under
$Ad(g)$, for all $g \in G$ , then any normal subsgroup $H$ in $G$ has
a dimension $0$ or $n = dim G$.

My soultion is here : assume $(m=)dimH \neq 0$. Then claim : $dimH=n$. My main thought is considering codimension of $H$.  Let $K \subset G$ be a codimesion of $G$. If $e_{1},.....,e_{m}$ is a basis of $H$, then $e_{m+1},...,e_{n}$ is a basis of $K$.  Now, put $Ad(g) : G \to End(H \oplus K) $ and $\psi:= Ad(g)|_{K}$. Since $G$ is irreducible, the invariant subspace $V \in T_{e}G$ such that $\psi(k)(v) \in V$ for all $k \in K$ does not exist except for trivial subspace 0. But, That's all I can do it.
I expect to logically induce $m=n$ from the above result, but I think that I am a wrong way to sort it out.
Another detour that I can imagine is the definition of the normal subgroup, left and right coset is coincident, yet I have not found strategy.
 A: First, I'll prove the hint in my comment:
The adjoint action of $G$ on its Lie algebra $\mathfrak{g}$ is defined as follows.  For each $g\in G$, let $C_g:G\rightarrow G$ be defined by $C_g(h) = ghg^{-1}$, that is, by conjugation.  The map $C_g$ is a Lie isomorphism with inverse $(C_g)^{-1} = C_{g^{-1}}$.
Then for the identity $e\in G$, $C_g(e) = geg^{-1} = e$, so $d_e(C_g):T_e G\rightarrow T_g G$.
The adjoint action is given by the map $G\rightarrow Gl(\mathfrak{g})$ with $g\mapsto d_e(C_g)$.
We claim that if $H\subseteq G$ is a normal subgroup, then this action preserves $T_e H\subseteq T_e G$.  To see this, let $v\in T_e H$ and let $\gamma:(-\epsilon,\epsilon)\rightarrow H$ be a smooth curve with $\gamma(0) = e$ and $\gamma'(0) = v$.
Because $\gamma(t)\in H$ for each $t$ in the domain, and because $H$ is normal in $G$, $C_g(\gamma(t)) = g\gamma(t) g^{-1}\in H$.  In other words, the curve $\alpha(t)=g \gamma(t) g^{-1}$ is a curve in $H$.  Moreover, $\alpha(0) = g\gamma(0)g^{-1} = geg^{-1} = e$, and $\alpha'(0)$ is, by definition, $d_e(C_g)(v)$.  Because $\alpha(t)\in H$ for all $t$ in the domain, $d_e(C_g)(v) = \alpha'(0)\in T_e H$.  This shows that $T_e H$ is preserved by the adjoint action of $G$.
Now, to actually answer the question in your post, if $H\subseteq G$ is normal, then $T_e H$ is an invariant subspace of $T_e G$.  Since the adjoint action is irreducible by assumption, it follows that $T_e H$ either has dimension $0$ or dimension $n$.
Now, if $\exp:\mathfrak{g}\rightarrow G$ is the group exponential map, then recall that $\exp$ is a diffeomorphism onto its image when restricted to a small enough open set $U\subseteq T_e G$.  It follows that $\exp|_{U\cap T_e H}:(U\cap T_e H)\rightarrow H$ is a diffeomorphism onto its image.  It's image is an open subset of $H^0$, the identity component of $H$, so it follows that $H^0$ has dimension $0$ or $n$.
Lastly, note that in a disconnected Lie group, all the components are diffeomorphic.  In fact, if $h$ is in some component of $H$, then left multiplication by $h$ is a diffeomorphsim which maps the identity component onto the component containing $h$.  In particular, all the components of $H$ have the same dimension, so that dimension is $0$ or $n$.
