# Is there a name for this homomorphism from $G$ to $\operatorname{Sym}([G:H])?$

Let $G$ be a group and $H$ its subgroup. Let $n=[G:H]$ be a cardinal number. Let $C=\{aH\,|\,a\in G\}.$ We have $n=\operatorname{card}(C).$ We define for any $g\in G$ the map $\phi_g:C\to C$ by the formula $$\phi_g(aH)=gaH.$$ The maps $\phi_g$ are well-defined because if $aH=bH,$ then $gaH=gbH$ for any $g\in G.$

$\phi_g$ must be 1-1 because if $gaH=gbH,$ then $$aH=g^{-1}(gaH)=g^{-1}(gbH)=bH.$$ $\phi_g$ must be onto because for $a,g\in G$ we have $$aH=g(g^{-1}aH)=\phi_g((g^{-1}a)H).$$

Therefore, $\phi_g$ is a permutation of $C.$ We can say that $$\{\phi_g\,|\,g\in G\}\subseteq \operatorname{Sym}(n).$$

Let $f:G\to\operatorname{Sym}(n)$ be defined by the formula

$$f(g)=\phi_g.$$

$f$ is a homomorphism because for $g_1,g_2\in G$ and $aH\in C$ we have

$$(f(g_1g_2))(aH)=\phi_{g_1g_2}(aH)=g_1g_2aH=(\phi_{g_1}\circ\phi_{g_2})(aH)=(f(g_1)\circ f(g_1))(aH).$$

I've just noticed this. Is there a name for this homomorphism? Are there names for its kernel and image?

Edit: Let's rename $f$ to $f_l$ and define $f_r$ analogously, but with right cosets instead of left cosets. Can it be for some $G$ and $H$ that

$$\ker f_l\neq \ker f_r?$$

Edit: OK, I think it can't be. We have

$$\begin{eqnarray} \ker f_l&=&\{g\in G \,|\, \phi_g=\operatorname{id}\}\\&=&\{g\in G\,|\,(\forall a\in G) gaH=aH\}\\&=&\{g\in G\,|\,(\forall a\in G) a^{-1}gaH=H\}\\&=&\{g\in G\,|\,(\forall a\in G) a^{-1}ga\in H\} \end{eqnarray}$$

Analogously,

$$\ker f_r = \{g\in G\,|\,(\forall a\in G) aga^{-1}\in H\}$$

and $$(\forall a\in G) a^{-1}ga\in H\iff (\forall a\in G) aga^{-1}\in H$$

So the kernels are equal.

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Nothing different happens when you take left/right cosets. And as for the terminology, I think there is no special name for this! Interestingly, there are a few results that I can think of depending on $n$. May be I'll write up an answer in case you'd like to know. This is a very important Homomorphism in Group Theory. –  user21436 Feb 11 '12 at 14:23
I would like to know! I will edit my last question to make it more precise. –  user23211 Feb 11 '12 at 14:26
It does not quite make sense to ask whether $f_l=f_r$ and $\operatorname{im} f_l= \operatorname{im}f_r$, because they act on different sets. –  Colin McQuillan Feb 11 '12 at 14:38
It looks fine. (One typo: $a^{-1}gaH=aH$ should be $a^{-1}gaH=H$) –  Colin McQuillan Feb 11 '12 at 15:30
It's somewhat more accurate to describe this as a map from G to $\operatorname{Sym}(G/H)$, the group of permutations of the coset space $G/H = \{aH:a\in G\}$. This latter group is isomorphic to $\operatorname{Sym([G:H])}$, but there isn't a canonical isomorphism between these groups -- it depends on choosing an ordering of $G/H$. –  Brad Feb 11 '12 at 16:20

$f$ is called a coset representation. The left and right coset reprsentations are isomorphic via the bijection sending $gH$ to $Hg^{-1}$.