Abstract Nonsense

Review of Group Theory: Cosets and Lagrange’s Theorem

Point of post: In this post we give a brief overview of the idea of  cosets, normal subgroups, inner automorphisms, etc. Remember, this is just a perfunctory run through to get us in the mood for our rep. theory talk.

Motivation

It turns out that just as was the case with cosets of vector spaces, cosets of groups (defined below) play an important role in our theory. We will then see a special case of how cosets of a subgroup carve up the ambient group in a nice way which will give us much information about when and when not a subgroup can be a group. In particular, we prove Lagrange’s theorem which states that (roughly) the order of a subgroup of a finite group must divide the order of the ambient group.

The Product of Subsets of a Group

In what follows it will be useful to define the product of two subsets of a group $G$. Explicitly if $\varnothing\subsetneq A,B\subseteq G$ then we define $AB$ to be the set

$AB=\left\{ab:a\in A\text{ and }b\in B\right\}$

If $A=\{a\}$ we denote $AB$ by $aB$ and similarly if $B=\{b\}$. Some basic theorems which can be derived about the product of subsets is

Theorem: Let $G$ be a group and $A,B,C\subseteq G$. Then,

\displaystyle \begin{aligned}&\mathbf{(1)}\quad (AB)C=A(BC)\\ &\mathbf{(2)}\quad A(B\cup C)=AB\cup BC\\ &\mathbf{(3)}\quad A(B\cap C)=AB\cap BC\\ &\mathbf{(4)}\quad \textit{If }A\leqslant G\textit{ and }B\subseteq A\textit{ then }AB=A\end{aligned}

Proof:

$\mathbf{(1)}$: This is clear since

\displaystyle \begin{aligned}A(BC) &= \left\{a(k):a\in A\text{ and }k\in BC\right\}\\ &= \left\{a(bc):a\in A\text{ and }b\in B\text{ and }c\in C\right\}\\ &= \left\{(ab)c:a\in A\text{ and }b\in B\text{ and }c\in C\right\}\\ &=\left\{kc:k\in AB\text{ and }c\in C\right\}\\ &= (AB)C\end{aligned}

$latex\ mathbf{(2)}$: Similarly,

\displaystyle \begin{aligned}A(B\cup C) &= \left\{ak:a\in A\text{ and }k\in B\cup C\right\}\\ &= \left\{ak:a\in A\text{ and }\left(k\in B\text{ or }k\in C\right)\right\}\\ &= \left\{ak:\left(a\in A\text{ and }k\in B\right)\text{ or }\left(a\in A\text{ and }k\in B\right)\right\}\\ &= \left\{ak:a\in A\text{ and }k\in B\right\}\cup\left\{ak:a\in A\text{ and }k\in C\right\}\\ &= AB\cup AC\end{aligned}

$\mathbf{(3)}$: Just as before

\displaystyle \begin{aligned}A\left(B\cap C\right) &= \left\{ak:a\in A\text{ and }\left(k\in B\text{ and }k\in C\right)\right\}\\ &= \left\{ak:\left(a\in A\text{ and }k\in B\right)\text{ and }\left(a\in A\text{ and }k\in C\right)\right\}\\ &= \left\{ak:a\in A\text{ and }k\in B\right\}\cap\left\{ak:a\in A\text{ and }k\in C\right\}\\ &= AB\cap AC\end{aligned}

$\mathbf{(4)}$: This follows since if $ab\in AB$ then $a\in A\text{ and }b\in B\subseteq A$ and since $A$ is closed under multiplication it follows that $ab\in A$. Conversely, if $a\in A$ and $b\in B$ then $ab^{-1}\in A$ and so  $a=(ab^{-1})b\in AB$. The conclusion follows.

$\blacksquare$

The last question we address is precisely when given two subgroups $H,K\leqslant G$ is $HK\leqslant G$?

Theorem: Let $G$ be a group and $H,K\leqslant G$. Then, $HK\leqslant G$ if and only if $HK=KH$.

Proof: Suppose first that $HK=KH$ and let $a,b\in HK$. Then, $a=h_1k_1$ and $b=h_2k_2$. Thus,

$ab^{-1}=h_1k_1k_2^{-1}h_2^{-1}$

But, note that if we label $k_3\overset{\text{def.}}{=}k_1k_2^{-1}$ then we may rewrite the above as

$h_1k_3h_2^{-1}$

But, since $k_3h2^{-1}\in HK$ and $HK=KH$ we know that $k_3h_2^{-1}=h_3k_4$ for some $h_3\in H$ and $k_4\in K$. Thus, we have that

$ab^{-1}=h_1h_3k_4$

but, if we label $h_4\overset{\text{def.}}{=}h_1h_3$ then we get that

$ab^{-1}=h_4k_4\in HK$

Thus, since $e\in HK$ we may conclude by previous theorem that $HK\leqslant G$.

Conversely, suppose that $HK\leqslant G$ and let $hk\in HK$ then since $HK\leqslant G$ we know that $(hk)^{-1}\in HK$ and thus $(hk)^{-1}=h'k'$ for some $h'\in H$ and $k'\in K$. Thus,

$hk=(h'k')^{-1}=(k')^{-1}h'^{-1}\in KH$

Conversely, we note that since $HK\leqslant G$ and $H,K\leqslant HK$ that $KH\subseteq HK$. The conclusion follows. $\blacksquare$

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

December 29, 2010 -

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