Minor Confusion Over Subgroup Index

Sep 2009
16
2
So, I'm reading in Algebra by Serge Lang and I have come to the definition that the index of H in G, (G:H) is the number of left cosets of H in G. I'm having trouble figuring this out.

Take for example,
G = Z (the set of integers)
H = 2Z

Now, I know that (G:H) = 2 from various examples, but I don't understand why. What are the two cosets of 2Z that form H?

By my definition, a coset of H is a subset of G of form aH where a is from G. But for any a in Z, a(2Z) = 2Z. So how do we get two cosets?

Thanks
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
So, I'm reading in Algebra by Serge Lang and I have come to the definition that the index of H in G, (G:H) is the number of left cosets of H in G. I'm having trouble figuring this out.

Take for example,
G = Z (the set of integers)
H = 2Z

Now, I know that (G:H) = 2 from various examples, but I don't understand why. What are the two cosets of 2Z that form H?

By my definition, a coset of H is a subset of G of form aH where a is from G. But for any a in Z, a(2Z) = 2Z. So how do we get two cosets?

Thanks
You always know given \(\displaystyle H\leqslant G\) that \(\displaystyle H\) itself will be one of the subgroups. Now, \(\displaystyle 2\mathbb{Z}+1=\left\{\cdots,-3,-1,1,3,\cdots\right\}\) and you're done. Why? Because \(\displaystyle 2\mathbb{Z}\cup \left(2\mathbb{Z}+1\right)=\mathbb{Z}\)
 
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Sep 2009
16
2
But is 2Z + 1 a coset of H?

That is the confusing part to me. I was under the impression that a coset of H had the form aH, not a + H.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
But is 2Z + 1 a coset of H?

That is the confusing part to me. I was under the impression that a coset of H had the form aH, not a + H.
Aha! Never fear...almost every person starting group theory has this trouble (myself included).

Really, that's bad notation but a necessary evil.

You'll remember that a group \(\displaystyle (G,*)\) is a non-empty set \(\displaystyle G\) along with an associative binary operation (function) \(\displaystyle *:G\times G\to G\) such that there is some \(\displaystyle e\in G\) with \(\displaystyle *(e,g)=*(g,e)=g\) for all \(\displaystyle g\in G\). For each \(\displaystyle g\in G\) there is some \(\displaystyle h\in G\) such that \(\displaystyle *(g,h)=*(h,g)=e\). And closure (but that is implicit in how we defined \(\displaystyle *\).

That said NO ONE wants to keep writing \(\displaystyle *\) everytime. So, instead of \(\displaystyle *(g,h)\) we write \(\displaystyle gh\). This does not mean multpilcation. This is just short hand notation for the function \(\displaystyle *\). Remember, \(\displaystyle *\) can be many, many things (function composition, multplication, matrix multiplication, addition, etc.). Now, you'll also remember that in general \(\displaystyle *(g,h)=gh\ne hg=*(h,g)\) and groups for which all the elements satisfy that aer called abelian. Now, it is customary (don't ask me why) to switch from using the short hand notation \(\displaystyle gh\) to \(\displaystyle g+h\) even though it means the same thing when the group is abelian.

So, it is true that a coset is of the form \(\displaystyle gH\) but that really means \(\displaystyle \left\{*(g,h):h\in H\right\}\) and for the group \(\displaystyle \mathbb{Z}\) we have that \(\displaystyle *\) means addition.
 
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Sep 2009
16
2
Ah, I see. Seems like I am constantly getting tripped up by notation in this whole group and ring theory business.

Thanks for your help!