LEC ALG2 7

Double cosets

Definition 96.1(Definition).

Let $H$ and $K$ be subgroups of $G$. Define

$$HgK=\{hgk|h\in H,k\in K\}.$$

The double cosets of $H$ and $K$ partition $G$. To see this, define an equivalence relation on $G$ by
$g\sim g^{\prime }$ if $g^{\prime }=hgk$ for some $h\in H$ and $k\in K$. It is easy to see that $\sim$ is reflexive and symmetric. If $g\sim g^{\prime }$ and $g^{\prime }\sim g^{\prime \prime }$, then $g=h^{\prime }{g}^{\prime }{k}^{\prime }$ and $g^{\prime }=h^{\prime \prime }{g}^{\prime \prime }{k}^{\prime \prime }$, which gives $g=h^{\prime }{h}^{\prime \prime }{g}^{\prime \prime }{k}^{\prime \prime }{k}^{\prime }$. Thus, $\sim$ is transitive. It follows that $\sim$ is an equivalence relation and therefore partitions $G$.

Remarks:

• If $K$ is normal, then $KgK=gKK=gK$.
• If $K$ is not normal, then $KgK$ contains a left coset and a right coset, namely $gK$ and $Kg$.

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Group actions

Definition 96.2(Definition).

Let $G$ be a group and let $X$ be a set. We say that $G$ acts on $X$ if we have a map $\varphi :G\times X\to X$ ($\varphi (g,x)$ is denoted by $gx$), called a group action, satisfying

1. $1_{G}x=x$ $\forall x\in X$, and
2. $g_1(g_{2}x)=(g_1{g}_2)x$ $\forall g_1,g_2\in G$ and $\forall x\in X$.

It can be notationally convenient to curry $\varphi$, so that it instead maps from $G$ to the set of maps $\{f:X\to X\}$, with each $g\in G$ being associated with a map ${\varphi }_g:X\to X$ satisfying

1. ${\varphi }_{1_G}$ is the identity on $X$, and

2. ${\varphi }_{g_1}{\varphi }_{g_2}={\varphi }_{g_1{g}_2}$.

It is easy to see that ${\varphi }_g$ is a bijection for all $g\in G$:

• $g{x}_1=g{x}_2$ $⟹$ ${\varphi }_g(x_1)={\varphi }_g(x_2)$ $⟹$ ${\varphi }_{g^{-1}}({\varphi }_g(x_1))={\varphi }_{g^{-1}}({\varphi }_g(x_2))$ $⟹$ ${\varphi }_{g^{-1}g}(x_1)={\varphi }_{g^{-1}g}(x_2)$ $⟹$ $x_1=x_2$.
• for all $x\in X$, $g(g^{-1}x)=x$, $i$.$e$, ${\varphi }_g(g^{-1}x)=x$.

Thus, the codomain of $\varphi$ can be reduced to the symmetric group $\text{Sym}(X)$, the group of all bijections from $X$ to $X$. This along with the defining properties of a group action makes $\varphi$ a group homomorphism.

Remark 96.3(Remark).

More generally, an action of a group $G$ on an object $A$ of a category $\text{C}$ is simply a homomorphism

$$\sigma :G\to {\text{Aut}}_{\text{C}}(A).$$

We will be exclusively dealing with the case $\text{C}=\text{Set}$, so ${\text{Aut}}_{\text{C}}(A)=\text{Sym}(A)$.

Remark 96.4(Aluffi (2009) Exercise III.9.3).

What we’ve defined above is a left group action. A right action would associate to each pair $(g,x)$ with and $x\in A$ an element $xg\in A$, such that $(x{g}_1)g_2=x(g_1{g}_2)$ for all $x\in X$, and $g_1,g_2\in G$. This is a different requirement than the one given above. Multiplication on the right in a group gives a prototypical example of a right action.

Every right action may be turned into a left action, as follows. If $G=(G,\cdot )$ is a group, define the ‘opposite group’ $G^{\circ }=(G,\bullet )$ supported on the same set $G$, by prescribing

$$(\forall g,h\in G):g\bullet h:=h\cdot g.$$

It is easy to verify that $G^{\circ }$ is indeed a group. The identity map $G\to G^{\circ }$ is an isomorphism iff $G$ is commutative. However, even if $G$ is not commutative, the map $G\to G^{\circ }$ given by $x\mapsto x^{-1}$ is an isomorphism, so $G\cong G^{\circ }$. Now, a right action of $G$ on a set $X$ is a map $\phi :G\to S_X$ such that ${\phi }_a\circ {\phi }_b={\phi }_{b\cdot a}$. However, since ${\phi }_{b\cdot a}={\phi }_{a\bullet b}$, $\phi$ is a homomorphism $G^{\circ }\to S_X$. ==Thus, giving a right action of $G$ on a set $X$ is the same as giving a left action of $G^{\circ }$ on $X$==. We therefore do not lose anything by restricting our study to left actions¹.

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If $\varphi$ is injective, then it is said to be faithful or effective.

Example 96.5(Example).

Let $G=S_4$ and $X=\{{\Pi }_1,{\Pi }_2,{\Pi }_3\}$, the latter as defined in Exm 92.9. Let $\varphi :G\times X\to X$ be defined by

$$\left(\tau ,\underset{=\{a,b\}\cup \{c,d\}}{{\Pi }_i}\right)\mapsto \tau {\Pi }_i\equiv \{\tau (a),\tau (b)\}\cup \{\tau (c),\tau (d)\}.$$

$\varphi$ satisfies the properties of a group action. Note that the curried version of $\varphi$ is not injective here, as $|\ker \varphi |=4$.

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Theorem 96.6(Cayley's theorem).

Any group $G$ of order $n$ is isomorphic to a subgroup of $S_n$.

Proof.

Let$G$ act on itself by left multiplication. That is, define $\varphi :G\to (G\to G)$ by $\varphi (g)={\varphi }_g$, ${\varphi }_g(g^{\prime })=g{g}^{\prime }$. Note that ${\varphi }_{1_G}$ is the identity on $G$, and ${\varphi }_{g_1}{\varphi }_{g_2}(g)=g_1{g}_{2}g={\varphi }_{g_1{g}_2}(g)$. Thus, $\varphi$ is a group action. If ${\varphi }_{g_1}={\varphi }_{g_2}$, then $g_{1}g=g_{2}g$ for all $g\in G$, which implies $g_1=g_2$, making $\varphi$ injective. Note $\varphi$ being a homomorphism makes $\mathrm{Im}\varphi$ is a subgroup of $\text{Sym}(G)$. So, ${\varphi }^{\prime }:G\to \mathrm{Im}\varphi$ defined by ${\varphi }^{\prime }(g)=\varphi (g)$ is a bijective homomorphism, or an isomorphism. Thus, $G\cong \mathrm{Im}\varphi <\text{Sym(G)}\cong S_n$.□

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Footnotes

1. the defining map $G\to S_X$ is a homomorphism for left group actions, unlike for right group actions, as seen here. The only reason for this is the manner in which we define function composition: $f\circ g$ is ‘first $g$ then $f$‘. ↩

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.