TT's Maths Thread

[kamil9876]

So suppose that and . Suppose we have written (product of m cycles). Then .

Thus:



So you see, has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation (I'm assuming u already know it is a group).

[TrueTears]

So suppose that and . Suppose we have written (product of m cycles). Then .

Thus:



So you see, has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation (I'm assuming u already know it is a group).

Thanks, I don't quite get what you mean by "...A_n is closed under conjugation (I'm assuming u already know it is a group). "? What is conjugation? and wat r u assuming to be a group..?

And also how have you shown A_n is a normal subgroup of S_n? Shouldn't you prove ? Why did you just pick an element from A_n....?

Also your last sentence doesn't seem to make sense lol

So you see, has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation

there should be a word b/w number and A_n?


What about the other Q btw?

[kamil9876]

it should be "is an even number, and so A_n is closed..."

Conjugation means taking conjugates, ie is a conjugate of . In order to show that a subgroup is normal it suffices to show that , that is what I meant by closed under conjugation. Of course you also have to show that it is a subgroup, but I'm assuming that u already know A_n is a subgroup.

(see here http://en.wikipedia.org/wiki/Normal_subgroup#Definitions, I used that definition of normal subgroup, there are other definitions such as yours but this one is the easiest to use when showing that something is normal)

[TrueTears]

Ahh okay thanks for that, never seen that definition before haha

Also for the 2nd question, i know that the first isomorphism theorem is that the image of is isomorphic to the quotient group

but i dono what to do with it, can u show me

oh nvm i got it lol

[TrueTears]

And just a final proof question: Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

Okay so what i started off with was let this element by g. Then we can create <g> and by lagrange's theorem we know that |G|/|<g>| = some constant. So (2n)/|<g>| = some constant, so how do i show that |<g>| is 2 and prove uniqueness?

Cheers

[kamil9876]

Well you can't really show that |<g>|=2 as g is arbitrary and so may be an element that is not of order 2. I guess this can be solved a number of ways depending on what theorems are available to you. Are we allowed to use the structure theorem for finitely generated abelian groups?

[TrueTears]

Are we allowed to use the structure theorem for finitely generated abelian groups?

wats that? i should know it, maybe i just dono the theorem name, what's the theorem about?

[kamil9876]

google it. I think you've seen it before. The fact that every finitely generated abelian group is a product of and quotients

[TrueTears]

ohhh yeah i know that, how does that play any role here though...?

[kamil9876]

So by the structure theorem we can assume that:



We know that it follows that one of the while all the others are odd, suppose and all others are odd. So



Can you now find an element of order 2 here? should be very easy.

Now try to prove uniqueness by proving that no other element is of order 2, how would such an element look like?

[TrueTears]

oh i think i get it now, an element of order 2 is (1, 0, 0, ..., 0)

now since all the other d_i's are odd, then any other element would have an order of at least 3, ie d_2 =>3, so say if d_2 was 3, then an element would look like (0, 1, 0, 0, ..., 0) which would have an order of at least 3.

[kamil9876]

In general, let p be a prime and n a number not having p is a factor. Then every abelian group of order pn has a unique element of order p.

[TrueTears]

alright, nice, that's a really nice result lol

[TrueTears]

Prove that for any group G, and any element g in G that gG = G

So if we pick g = e, then eG = G, but the question says g is ANY element of G, how can i show that gG = G for some g not the identity?

[/0]

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Let , then for all by closure. Then, since was arbitrary, .

:
For any , does there exist such that ? By construction, yes: . Hence, for every element in , that element exists in , so .