Let $G$ be a group and $V_j$ , $j =1,2$ be irreducible representations of $G$. Show that any map $\phi: V_1 \rightarrow V_2$ of representations is either an isomorphism or zero. The Hint I got was: Show that $Im(\phi)$ and $Ker(\phi)$ are Sub-representations and use the defintiion of Irreducibility of $V_{1,2}$.

I think what I have to do is the following: If we show that the kernel and the image of $\phi$ are subrepresentations of $G$ we can deduce information about $\phi$ from the fact that the only "allowed" subrepresentations are $V_{1,2}$ and $\left\{0\right\}$ because of the irreducability of $V_j$

So now I'm trying to show that $ker (\phi)$ is a subrepresentation of $V_1$:

$ker(\phi) = \left\{v \in V_1 | \phi(v) = \phi (\rho_1(g))(v) = \rho_2(g)(\phi(v)) = {id}_{V_2}\right\}$ where $\rho_1$ is the representation of $G \rightarrow GL(V_1)$ and $\rho_2$ is the representation of $G \rightarrow GL(V_2)$. Now a if I want to show that $ker(\phi)$ is a subrepresentation I need to show that the group action $g * ker(\phi)$ maps $ker(\phi) \rightarrow ker(\phi)$. But the elements of $ker(\phi)$ are vectors from the vector space $V_1$, so how can I multiply these with an element from $G$?

I think there's something where I missed the point sorta completely so if someone could point that out for me I'd appreciate it!

Cheers in advance!

  • $\begingroup$ I have never seen this use of the term "representation" before. Is this what you learnt in your class? It is more usual to define a group representation to be a group homomorphism from the group $G$ to the automorphism group ${\rm GL}(V)$ of a vector space $V$. Then $V$ itself is referred to as a module with a group action. So what you are calling representations and subrepresentations would usually be called modules and submodules. $\endgroup$
    – Derek Holt
    Mar 17, 2013 at 13:18
  • $\begingroup$ Yeah that's basically just our notation. If we say $V$ is a representation of $G$ then we mean that there is a map $\rho : G \rightarrow GL(V)$ $\endgroup$
    – Howdy Ho
    Mar 17, 2013 at 13:22

1 Answer 1


$g \cdot v_1$ is just a shorthand way of saying $\rho_1(g)(v_1)$. That is each, since $\rho_1 : G \rightarrow GL(V_1)$ is a homomorphism, each $g \in G$ gives rise to an invertible linear map $\rho_1(g) : V_1 \rightarrow V_1$, and $g \cdot v_1$ is a notationally convenient way of saying 'the vector in $V_1$ that is the image of $v_1$ under the linear map $\rho_1(g)$.

  • $\begingroup$ I see! So basically I have to show that $\rho_1(g)(ker(\phi))$ maps strictly from the kernel to the kernel? $\endgroup$
    – Howdy Ho
    Mar 17, 2013 at 13:20
  • $\begingroup$ OH! No! I have to show that for $v \in ker(\phi)$, $\rho_1(g)(v) \in ker(\phi)$ i.e: $\phi((\rho_1(g)(v)) \in ker(\phi)$ for all $v \in ker(\phi)$.. right? $\endgroup$
    – Howdy Ho
    Mar 17, 2013 at 13:45
  • $\begingroup$ Here's how the notation comes in handy: to say $\phi$ is a map of representations is equivalent to $\phi(g \cdot v_1) = g \cdot \phi(v_1)$. So now showing $v_1 \in \ker(\phi) \Rightarrow g \cdot v_1 \in \ker(\phi)$ is just a formality. $\endgroup$ Mar 17, 2013 at 15:26
  • $\begingroup$ Yep, I was actually unsure if what I did was correct, I wrote down the definition and.. there was the correct result, haha. Thanks anyway for your help! $\endgroup$
    – Howdy Ho
    Mar 17, 2013 at 16:01

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