Question

2nd Year Group Theory

Answer 1

I believe I can write it informally, I am just having issues with formalities. Does it go something like:

Since <S,T> is the smallest subgroup of G such that all elements of S and T are contained within the subgroup, <<S>,T> is the smallest subgroup of G such that all elements of <S> and T are contained within the subgroup. Furthermore, <S> is the smallest subgroup of G such that all elements of S are contained within it. So can this be looked at as <S> with additional elements T added to it?

Answer 2

Sort of yes, but to do this carefully, write out the definition of
<S,T>
I would then aim to show that
\[ <S,T> \ \subset \ <<S>,T> \ \subset \ <S,<T>> \ \subset \ <<S>,<T>> \ \subset \ <S,T> \]
and hence must all be equal

Answer 3

\[ <S,T> \subset <<S>,T> \subset <S,<T>> \subset <<S>,<T>> \subset <S,T> \]

Answer 4

just to clarify, above does the \[\subset mean \subseteq ?\]

Answer 5

The definition I was using was the wordy sentence I wrote out repeatedly above, is there a more concise mathematical definition?

Answer 6

Yes, the subset symbol usually also means "...or equal to".

In terms of mathematical definition, for any subset A of G, the subgroup generated by A--denoted by <A>--is the smallest subgroup of G containing all elements of A.

By extension, <S,T> is the smallest subgroup of G containing both S and T.

Answer 7

Okay thanks for the clarification, I understand that. I have had a particularly long day and I am struggling to logically get through this problem, can you shunt me in the right direction?

Answer 8

The first step is obvious, I think.

Suppose G1 = <S,T>. Then as <S> contains S, it must certainly be the case
\[ G_1 = <S,T> \ \subset \ <<S>,T> = G_2 \]
because if that were not the case and there was an element g in G1 not in G2, then as
S \subset <S>, g need not be a member of G1. Contradiction.

***

In terms of overall strategy, it might be easier to show that
Step 1: <S,T> subset <<S>,T> subset <<S>,<T>> (easy)
and
Step 2: <S,T> subset <S,<T>> subset <<S>,<T>> (easy)
and
Step 3: <<S>,<T>> subset <S,T> (slightly trickier)

Answer 9

I think I was over complicating it in my head.

Answer 10

How did you show Step 3: <<S>,<T>> subset <S,T> ???

Answer 11

yo james, can you come to my problem next?
http://openstudy.com/study#/updates/511179fae4b0d9aa3c487dfb

Answer 12

I would argue by contradiction

Answer 13

Assuming statement in step 3 is not true, what do we contradict?