# On permutations of left cosets

Let $K$ be a subgroup of some group $H$; let $X$ be the set of left cosets of $K$, i.e. $X = \{hK: h \in H\}$; and let $G$ be the group of permutations of $X$. For all $h \in H$, let $f\,(h) \in G$ be the permutation of $X$ that sends every coset $h'K$ to the coset $hh'K$. It's easy to see that the map $f:H\to G$ is a homomorphism of groups. My question is

what is the kernel of $f\;\;$?

(I understand that $K\subseteq \ker(f\,)$, and also that if $K$ is contained in the center of $H$, then $\ker(f\,) = K$, but I'm interested in the general case.)

Thanks!

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For each $h\in H$, the map $f\,(h):X\to X$ is a permutation of $X$, because it is both injective $(f\,(h)(h'K) =$ $hh'K =$ $hh''K =$ $f\,(h)(h''K)$ iff $h'K = h''K$) and surjective ($\forall \; h'K\in X$, $f\,(h)(h^{-1}h'K) =$ $hh^{-1}h'K =$ $h'K$). The map $f:H\to G$ is a homomorphism of groups, since $\forall h, h' \in H,$ $\forall h''K\in X$, $f\,(hh')(h''K) =$ $hh'h''K =$ $f\,(h)(h'h''K) =$ $f\,(h)(f\,(h')(h''K))$, so $f\,(hh') = f\,(h)\circ f\,(h')$, and also, $\forall \, hK \in X$, $f\,(1_H)(hK) =$ $1_H\,hK =$ $hK$ so $f\,(1_H) =$ $1_G$ . –  kjo Dec 26 '11 at 13:37

It is not true in general that $K$ is contained in the kernel of $f$. The kernel contains $K$ if and only if it equals $K$, because

$Ker(f) = \{h \in H: haK = aK \text{ for every } a \in H \} = \bigcap_{a \in H} a^{-1}Ka$

and so $Ker(f) \leq K$.

For example, let $H = S_3$ and $K = \{(1), (12)\}$. Then $K$ has cosets $C_1 = \{(1), (12)\}$, $C_2 = \{(13),(123)\}$ and $C_3 = \{(23), (132)\}$. Now the permutation $f((12))$ maps $C_2$ to $C_3$, so it is not the identity. Thus $(12) \notin Ker(f)$. In particular, $K$ is not contained in $Ker(f)$.

So although $kK = K$ for every $k \in K$, it is not necessarily true that $kaK = aK$ for arbitrary $a \in H$. But it is true if (and only if) $K$ is normal in $H$.

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That's nice, thanks! –  kjo Dec 26 '11 at 14:05

After seeing m.k.'s proof, I realized that

$$h \in \ker(f\,) \Leftrightarrow h1_HK = 1_HK \Leftrightarrow hK = K \Leftrightarrow h \in K.$$

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$h1_HK = 1_H K$ does not necessarily imply $h \in Ker(f)$, so this is not correct. See the example in my answer. The kernel does not always equal $K$! –  Mikko Korhonen Dec 26 '11 at 14:35
@m.k.I think our edits crossed, because did not see the point you make about how $K$ is not necessarily a subset of $\ker(f\,)$. Thanks for the correction. –  kjo Dec 26 '11 at 16:22
Just to add to m.k.'s response, the $Ker(f)=\bigcap_{a \in H} a^{-1}Ka=\text{core}_H(K)$, called the normal core of $K$ in $H$, and is the largest normal subgroup of H that is contained in K, see this for more information.
I am curious though kjo why you are asking this particular question, because I did some work related to your question. If the group $H$ is the symmetry group of some set of objects acting transitively, one can talk about a coloring of the objects by assigning a unique color to each left coset of $K$. If I am not mistaken, $G$ is in fact, just $H$, called in this setting the color symmetry group of the coloring, and the $ker(f)$ is called the colored fixing group.
My interest in the question is more idle curiosity than anything else; I was just trying to fill in an omitted step in a proof. To do this, I needed a clearer sense of the relationship between $K$ and $\ker(f\,)$. –  kjo Dec 26 '11 at 16:28