[proofplan]
Pick a non-zero vector $v \in V$ and define a linear map $\theta: \mathbb{F}G \to V$ by sending each basis vector $e_g$ to $gv$ and extending linearly. Verify that $\theta$ is a $G$-homomorphism and that $\operatorname{im}\theta$ is a non-zero $G$-subspace of $V$ — irreducibility of $V$ then forces $\operatorname{im}\theta = V$. By Maschke's theorem (which applies because $G$ is finite and $\operatorname{char}\mathbb{F} = 0$), $\ker \theta$ has a $G$-invariant complement $W \leq \mathbb{F}G$. The first isomorphism theorem then identifies $W$ with $V$ as $G$-spaces, exhibiting $W$ as a copy of $V$ inside $\mathbb{F}G$.
[/proofplan]
[step:Choose a non-zero vector and define the candidate $G$-homomorphism]
Since $V \neq 0$ (irreducible representations are non-zero by convention), pick any $v \in V$ with $v \neq 0$. Recall that the group algebra $\mathbb{F}G$ has $\mathbb{F}$-basis $\{e_g : g \in G\}$, and the regular representation $\rho_{\mathrm{reg}}: G \to \operatorname{GL}(\mathbb{F}G)$ is determined by $h \cdot e_g = e_{hg}$ for $h, g \in G$, extended linearly. Define
\begin{align*}
\theta: \mathbb{F}G &\to V \\
\sum_{g \in G} a_g e_g &\mapsto \sum_{g \in G} a_g\, (gv).
\end{align*}
This is the unique $\mathbb{F}$-linear map sending $e_g \mapsto gv$ for each $g \in G$.
[/step]
[step:Verify that $\theta$ is a $G$-homomorphism]
We must check $\theta(h \cdot x) = h \cdot \theta(x)$ for all $h \in G$ and $x \in \mathbb{F}G$. By $\mathbb{F}$-linearity of both sides in $x$, it suffices to check the identity on the basis vectors $e_g$.
Fix $h, g \in G$. On the one hand,
\begin{align*}
\theta(h \cdot e_g) = \theta(e_{hg}) = (hg)v,
\end{align*}
using the definition of the regular action and of $\theta$. On the other,
\begin{align*}
h \cdot \theta(e_g) = h \cdot (gv) = (hg)v,
\end{align*}
using the definition of $\theta$ and the homomorphism property $\rho(h)\rho(g) = \rho(hg)$ of the representation $\rho$. The two expressions agree, so $\theta(h \cdot e_g) = h \cdot \theta(e_g)$ for every $h, g \in G$. By linearity, $\theta$ is a $G$-homomorphism.
[/step]
[step:Show $\operatorname{im}\theta = V$]
The image $\operatorname{im}\theta$ is a $G$-subspace of $V$: it is an $\mathbb{F}$-subspace since $\theta$ is $\mathbb{F}$-linear, and $G$-invariant since $\theta$ is a $G$-homomorphism (for any $h \in G$ and $y = \theta(x) \in \operatorname{im}\theta$, we have $hy = h\theta(x) = \theta(hx) \in \operatorname{im}\theta$).
The image is non-zero: $\theta(e_{1_G}) = 1_G \cdot v = v \neq 0$.
Since $V$ is irreducible, its only $G$-subspaces are $0$ and $V$. Combined with $\operatorname{im}\theta \neq 0$, we conclude $\operatorname{im}\theta = V$.
[/step]
[step:Apply Maschke's theorem to obtain a $G$-invariant complement of $\ker\theta$ in $\mathbb{F}G$]
The kernel $\ker\theta$ is a $G$-subspace of $\mathbb{F}G$: it is an $\mathbb{F}$-subspace since $\theta$ is linear, and $G$-invariant since $\theta$ is a $G$-homomorphism (for any $h \in G$ and $x \in \ker\theta$, $\theta(hx) = h\theta(x) = h \cdot 0 = 0$, so $hx \in \ker\theta$).
We invoke [Maschke's Theorem](/theorems/2409). Maschke's theorem requires (i) $G$ finite and (ii) the characteristic of the base field does not divide $|G|$. Hypothesis (i) is given. For (ii), the field has characteristic $0$, so $\operatorname{char}\mathbb{F} \nmid |G|$ holds vacuously. Maschke's theorem applies and supplies a $G$-subspace $W \leq \mathbb{F}G$ such that
\begin{align*}
\mathbb{F}G = \ker\theta \oplus W.
\end{align*}
[/step]
[step:Identify $W$ with $V$ as $G$-spaces via the first isomorphism theorem]
Consider the restriction $\theta|_W: W \to V$. This is a $G$-homomorphism (restriction of a $G$-homomorphism to a $G$-subspace).
We claim $\theta|_W$ is injective. Suppose $x \in W$ and $\theta(x) = 0$. Then $x \in \ker\theta \cap W$. The direct-sum decomposition $\mathbb{F}G = \ker\theta \oplus W$ states that $\ker\theta \cap W = 0$, so $x = 0$.
We claim $\theta|_W$ is surjective. Given $y \in V$, by $\operatorname{im}\theta = V$ there is some $x \in \mathbb{F}G$ with $\theta(x) = y$. Decompose $x = x_1 + x_2$ with $x_1 \in \ker\theta$ and $x_2 \in W$. Then
\begin{align*}
\theta(x_2) = \theta(x) - \theta(x_1) = y - 0 = y.
\end{align*}
So $\theta|_W$ is surjective onto $V$.
Therefore $\theta|_W: W \to V$ is a $G$-isomorphism, and $W$ is a copy of $V$ inside $\mathbb{F}G = \rho_{\mathrm{reg}}$, completing the proof.
[guided]
The aim is to find a $G$-subspace of $\mathbb{F}G$ isomorphic to $V$. The strategy splits into two halves.
**Half 1: produce a $G$-homomorphism $\mathbb{F}G \to V$ with image $V$.** Any $G$-homomorphism out of $\mathbb{F}G$ is determined by where it sends the identity element's basis vector $e_{1_G}$, because $e_g = g \cdot e_{1_G}$ in $\mathbb{F}G$. Picking $e_{1_G} \mapsto v$ for some non-zero $v \in V$ extends uniquely to the $G$-homomorphism $\theta$ defined above. Why is the image all of $V$? Because (a) the image contains $v \neq 0$, so it is a non-zero $G$-subspace of $V$, and (b) $V$ is irreducible, so the only non-zero $G$-subspace of $V$ is $V$ itself. The image is forced to be all of $V$.
**Half 2: lift $V$ from a quotient of $\mathbb{F}G$ to a subspace of $\mathbb{F}G$.** The first isomorphism theorem gives $\mathbb{F}G / \ker\theta \cong V$ as $G$-spaces (the quotient is well-defined as a $G$-space because $\ker\theta$ is a $G$-subspace). But we want $V$ as a sub-representation, not a quotient. Maschke's theorem is exactly the tool that converts quotients to direct summands: it says every $G$-subspace of a finite-dimensional $G$-space (over a field of suitable characteristic) admits a $G$-invariant complement. We verify Maschke's hypotheses: $G$ is finite (given) and $\operatorname{char}\mathbb{F} = 0$ does not divide $|G|$ (any positive integer is non-zero in characteristic $0$). Maschke applies, giving $\mathbb{F}G = \ker\theta \oplus W$ for some $G$-subspace $W$.
The restriction $\theta|_W: W \to V$ is the candidate isomorphism. It is injective because $\ker(\theta|_W) = \ker\theta \cap W = 0$ (intersection of complementary direct summands is zero), and surjective because every $y \in V = \operatorname{im}\theta$ has some preimage in $\mathbb{F}G$, and we can adjust the preimage by an element of $\ker\theta$ to land it in $W$.
**What goes wrong without Maschke?** In positive characteristic $p \mid |G|$, Maschke fails — $\ker\theta$ may have no $G$-invariant complement. Then $\mathbb{F}G / \ker\theta \cong V$ as $G$-spaces, but no copy of $V$ need sit inside $\mathbb{F}G$ as a sub-representation. The result of this theorem is therefore **false** in modular representation theory; the characteristic-$0$ hypothesis is essential.
**What if $V$ were reducible?** The image of $\theta$ would still be a non-zero $G$-subspace of $V$, but it might be a **proper** subspace, and we would not get a copy of all of $V$ inside $\mathbb{F}G$ via this construction. Irreducibility of $V$ is what forces $\operatorname{im}\theta = V$.
[/guided]
[/step]
