Let $Y: G/N→ GL_d(\mathbb{C})$ be a representation defined by $Y(gN) = X(g)$. Where $N = {\{g\in G: X(g) = I}\}$ is the kernel.
Prove that $Y$ is irreducible if and only if $X$ is.
attempt: A nonzero representation $V$ of $G$ is said to be irreducible if its only subrepresentation are zero and V itself.
let $Y: G/N→ GL_d(\mathbb{C})$.
Suppose $X$ is irreducible , can someone please help me with the converse? I am stuck on that one. Thank you!
It's sometimes useful to think of representations in the following way. Let $V$ be the complex vector space $\mathbb{C}^d$. A representation of $G$ is a group homomorphism $X: G \rightarrow GL_d(\mathbb{C})$, right? So, for each $g \in G$, $X(g)$ is an invertible matrix. But invertible matrices $A$ correspond to linear transformations $T_A:V \rightarrow V$ which are invertible.
A vector subspace of $V$ is a nonempty subset $W$ of $V$ which is closed under scalar multiplication and addition. A vector subspace $W$ is $G$-invariant (with respect to a given representation $X: G \rightarrow GL_d(\mathbb{C})$) if for all $g \in G$, and all $w \in W$, the element $T_{X(g)}(w)$ still lies in $W$.
To say that a representation $X: G \rightarrow GL_d(\mathbb{C})$ of a group $G$ is irreducible is to say that the only $G$-invariant subspaces of $V$ are the zero subspace and $V$ itself.
Now the induced representation $Y: G/N \rightarrow GL_d(\mathbb{C})$ of $G/N$ is basically the 'same' as the representation of $G$, except we are now considering the inputs to be cosets (special subsets) of $G$ instead of elements of $G$. In other words, $Y(gN) = X(g)$. So, it is intuitively obvious (and you should carefully prove) that a subspace $W$ of $V$ is $G$-invariant (with respect to the representation $X$) if and only if it is $G/N$-invariant (with respect to the representation $Y$).
Hence $X$ is irreducible if and only if $Y$ is irreducible.