# 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 -

## 12 Comments »

1. […] Point of post: This is a continuation of this post. […]

Pingback by Review of Group Theory: Cosets and Lagrange’s Theorem (Pt. II) « Abstract Nonsense | December 29, 2010 | Reply

2. […] it should be noted that all of the above notation is loose notation for the product of subsets). Conversely, suppose that then […]

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3. […] Recall that it suffices to prove that .  Thus, we notice […]

Pingback by Review of Group Theory: The Second Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

4. […] By a previous theorem it suffices to show that . To do this we merely note that by definition for each one has and thus […]

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5. […] 2: Let be a homomorphism.  Suppose there are maps and such that (where is the inner autormorphism induced by in ). Then, […]

Pingback by Groups of Order pq (Pt. II) « Abstract Nonsense | April 19, 2011 | Reply

6. […] is the canonical projection. Evidently is a vector space over and evidently every group endomorphism will be a linear endomorphism. Thus, since by definition is by definition where is the ordered basis we have from basic linear algebra that and is a monomorphism. Thus and so by Lagrange’s theorem . […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) ( January 2003)) « Abstract Nonsense | May 1, 2011 | Reply

7. […] is a natural extension of the notion of cosets which are double cosets, namely cosets on ‘both’ sides. While of algebraic interest in […]

Pingback by Double Cosets « Abstract Nonsense | May 2, 2011 | Reply

8. […] where we have evidently used Lagrange’s theorem. […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (January-2004) « Abstract Nonsense | May 6, 2011 | Reply

9. […] from Lagrange’s theorem we have that if is a -group then every element of has prime power order. Conversely, evidently […]

Pingback by Actions by p-Groups (Pt. II) « Abstract Nonsense | September 15, 2011 | Reply

10. […] from Lagrange’s theorem we have that if is a -group then every element of has prime power order. Conversely, evidently […]

Pingback by Actions by p-Groups (Pt. II) « Abstract Nonsense | September 15, 2011 | Reply

11. […] implied by the previous property since if and is the cyclic subgroup generated by then by Lagrange’s Theorem for each and so, in particular constitutes solutions of and so clearly there can’t be […]

Pingback by Unit Group of a Finite Field is Cyclic « Abstract Nonsense | September 21, 2011 | Reply

12. […] order dividing . Suppose first that , then clearly every element of has order dividing since (by Lagrange’s theorem, of course) since any such elements order would divide which divides . Thus, if then . Suppose […]

Pingback by Homomorphisms Between Finitely Generated Abelian Groups (Pt. I) « Abstract Nonsense | November 14, 2011 | Reply