Let $\phi: G → GL_d(\mathbb{C})$ be a homomorphism. And let the kernel be $N = {\{g\in G: \phi(g) = I}\}$, and $N $ is a normal subgroup of $G$.

Show that for the coset representation , $N = \cap_ig_iHg_i^{-1}$, where the $g_i$ are the transversal.

Proof: Suppose $N$ is the kernel of $\phi$. Then $N = gNg^{-1}$ for all $g\in G$. And let $H$ be a subgroup of $G$.

Then $\cap_ig_iHg_i^{-1} = g_1Hg^{-1}\cap g_2Hg_2^{-1}.....\cap g_kHg_k^{-1}$.

So recall $gg_iH = g_iH$.

Can someone please help?

  • $\begingroup$ What is $H$ ? Is it a normal subgroup ? $\endgroup$ Feb 29, 2016 at 22:07
  • $\begingroup$ It does not say. I dont assume it is $\endgroup$
    – user824284
    Feb 29, 2016 at 23:27

1 Answer 1


Let $G$ be a group, and $H$ a subgroup. Then $G$ acts on $G/H = \{gH|g\in G\}$ by left translation. You seek the kernel of this action (your question uses the language of linear representations, but this is overkill, since the kernel of an action is the same as the kernel of its associated linear representation).

Now let $a\in G$ be in this kernel, meaning that $agH = gH$ for all $g\in G$ (or just for $g$ running over a set of representatives, if you will). Then $g^{-1}agH=H$, so $g^{-1}ag \in H$, which is equivalent to $a\in gHg^{-1}$. So indeed $a\in \bigcap gHg^{-1}$.

The converse is even easier (and you can actually reason directly through equivalences, though I wouldn't advise to do so if you're uneasy with the material).


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