element of order II, in 2n order group

let G be a group with even order, G has an element of order 2 or it is own inverse

I know two ways to prove that but how about this ?

my work

let if we are done if not let

by Lagrange's theorem

let this an subgroup of G with even order

let if then there exist an element in K say k such that in G

if not then let it is clear that

and we remove two divisors from G order

if we repeat this operation in the end we will get an element of order 2

is my proof good

as i said I do not want an answer just check mine

Thanks in advance

Re: element of order II, in 2n order group

You are assuming that is normal in , otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...(the easiest example I can think of is - the elements of order here do not generate normal subgroups. An example where the non-normal subgroups are of order not 2 would be in - any element of order 3 will not generate a normal subgroup. It will generate a subgroup, but it won't be normal...)

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Swlabr** You are assuming that

is normal in

, otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...

but is still a subgroup and it is order divides G order

and which is still even since H order is not even

Re: element of order II, in 2n order group

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Swlabr** No. G/H denotes the SET of cosets

. They form a group IF AND ONLY IF H is normal in G.

For example, let

. Then your cosets are

,

and

. They do NOT form a group - can you work out why not?

I know it is the left costs, but i forgot(it was known) that to be a subgroup it should be normal

if we change the representative give another result

Quote:

Originally Posted by

**Swlabr** , if you take the group of the integers under addition, then quotient out the subgroup

which is normal because the integers form an abelian group, then your quotient is cyclic of order 2 (why?). The integers contain no element of order 2, so your quotient group is not a subgroup.

in my proof i was talking about a group of finite order

which is of order 2 any group of prime order is cyclic

Thanks very much

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Amer** in my proof i was talking about a group of finite order

which is of order 2 any group of prime order is cyclic

Thanks very much

This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head. Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group which has a normal subgroup such that is not isomorphic to any subgroup of ...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Swlabr** This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head. Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group

which has a normal subgroup

such that

is not isomorphic to any subgroup of

...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!

thanks very much you helped me alot sure I will rethink about my proof

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Swlabr** This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head.

** Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H** (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group

which has a normal subgroup

such that

is not isomorphic to any subgroup of

...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!

i think an example of such a group would be Q8. Q8/Z(Q8) is of order 4, and is isomorphic to the klein 4-group, but any subgroup of Q8 of order 4 is cyclic. and in point of fact, Q8 cannot be realized as a semi-direct product of subgroups; even though Q8/<i> (for example) is isomorphic to {1,-1}, we have {1,-1} ∩ <i> = {1,-1}, so Q8 doesn't "split" over <i> (a similar logic holds for <j> and <k>). it also doesn't "split" over Z(Q8), for essentially the same reason.

(as this shows G/H can be isomorphic to a subgroup of G even if we don't have a semi-direct product).

of course, 2 is prime, so the result Amer wants to prove is just a simple example of the sylow theorems, or (even more pointedly) of cauchy's theorem for finite groups.

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Deveno** i think an example of such a group would be Q8. Q8/Z(Q8) is of order 4, and is isomorphic to the klein 4-group, but any subgroup of Q8 of order 4 is cyclic. and in point of fact, Q8 cannot be realized as a semi-direct product of subgroups; even though Q8/<i> (for example) is isomorphic to {1,-1}, we have {1,-1} ∩ <i> = {1,-1}, so Q8 doesn't "split" over <i> (a similar logic holds for <j> and <k>). it also doesn't "split" over Z(Q8), for essentially the same reason.

(as this shows G/H can be isomorphic to a subgroup of G even if we don't have a semi-direct product).

of course, 2 is prime, so the result Amer wants to prove is just a simple example of the sylow theorems, or (even more pointedly) of cauchy's theorem for finite groups.

Gah - I was thinking of the Quaternion group, and then I thought that i and -1 formed the Klein 4-group. Not entirely sure why I thought that...

Re: element of order II, in 2n order group

Quote:

Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H

any example please or a website contains a proof for that statement

Thanks

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Amer** any example please or a website contains a proof for that statement

Thanks

Emm...semi-direct products are complicated. Look them up on wikipedia, or you could try looking up your favourite group-theory text. I'm not sure if they'd be in an introductory text though...they're a generalisation of direct products, but where only one of the subgroups is normal...

Re: element of order II, in 2n order group

Quote:

Originally Posted by

**Swlabr** Emm...semi-direct products are complicated. Look them up on wikipedia, or you could try looking up your favourite group-theory text. I'm not sure if they'd be in an introductory text though...they're a generalisation of direct products, but where only one of the subgroups is normal...

it is ok I will search for it never mind

I read it in Wiki I figured it thanks

Re: element of order II, in 2n order group

if you know about group actions, semi-direct products (at least internal ones) aren't that scary. suppose we have a group G, with a normal subgroup N. suppose H is another subgroup of G (not necessarily normal). we can let H act on N, by defining . since N is normal, this is a perfectly good action of H on N.

now if, furthermore, we have N∩H = {e}, and G = NH, we say G is the (internal) semi-direct product of N and H. by the 2nd isomorphism theorem:

, since the intersection of H and N is trivial.

for an EXTERNAL semi-direct product, we want to create a "parent group" for 2 groups H and N, so that they act just like the internal semi-direct product above. the trouble is, we don't have a multiplication to say how H and N will interact. fortunately, there is a way around this. note that h induces an inner automorphism in the internal direct product. so the idea is, if we have a homomorphism:

, we can try to build the "product" of H and N so that H acts on N by conjugation.

well, we need a set that has elements, so we take the cartesian product to be our underlying set.

now, here's where it gets messy. instead of using the usual "component-wise" multiplication, we instead let H "twist" N, before multiplying the components. formally:

where is the automorphism of N induced by , that is, the image .

now, it's not immediately clear we've done what we set out to do. first of all, it's not obvious that this product is even associative. so we compute:

(whew!)

well, a group needs an identity, so we'd better have one. the obvious candidate is . so let's check:

but the identity of H has to induce the identity automorphism of N, so:

so we at least have a monoid. do we have inverses? given (n,h) consider the element . again, we compute:

, so we indeed have a group.

(continued on next post)

Re: element of order II, in 2n order group

is this the group we set out to create? that is, does H truly act on N by conjugation?

well, let's unravel this. first let's create two new subgroups of our external semi-direct product (which we write as , to emphasize the fact that we are using the homomorphism ). so we define:

.

first we want to show that N' is normal in G. so:

, which is clearly in N'.

it's also clear that , and by construction, G = N'H'.

so the only thing left to show is that H' acts on N' by conjugation.

but what does mean in this context? we have to actually use H' and N', so we have:

.

so we see that conjugating by really does give us the action:

.

well!

as you can see, explicitly calculating (external) semi-direct products, can be an arduous process. but there are some simplifications. first and foremost, it might be that the homomorphism , could be a trivial one. in this case, every automorphism induced by every element of h, is just the identity automorphism of N, in which case we get the multiplication:

which is just a DIRECT product. but there are some interesting non-trivial semi-direct products.

suppose that we have the cyclic group , and the cyclic group . besides the trivial homomorphism:

for k = 0,1, we also have the following homomorphism:

, where inv is the automorphism that sends .

in the latter case has the following multiplication:

where the first coordinate is mod n, and the second coordinate is mod 2.

this group is generated by (1,0) and (0,1) and furthermore:

(1,0)(0,1) = (1,1) = (0,1)(-1,0).

if we map (1,0)--->r and (0,1)-->s this says that , which is just our old friend in disguise. this reflects the fact (ooh, a pun), that multiplying by a reflection "flips" or "twists" the n-gon we are considering.

Re: element of order II, in 2n order group

Thanks very much for your posts I am studying it, and if i have a question i will inform you