Abstract Nonsense

Crushing one theorem at a time

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


Advertisements

December 29, 2010 - Posted by | Algebra, Group Theory, Uncategorized | , , ,

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 […]

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

  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 […]

    Pingback by Review of Group Theory: Sylow’s Theorems (Pt. II) « Abstract Nonsense | January 8, 2011 | Reply

  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


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: