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 $\displaystyle x \in G $ if $\displaystyle |x| = 2 $ we are done if not let

$\displaystyle H = < x > $

by Lagrange's theorem $\displaystyle \[G:H] \mid | G | $

let $\displaystyle K = G/H $ this an subgroup of G with even order

let $\displaystyle y \in K $ if $\displaystyle | y | = 2 $ then there exist an element in K say k such that $\displaystyle |yk| = 2 $ in G

if not then let $\displaystyle M = K / <y> $ it is clear that $\displaystyle |M| \mid |G| $

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 $\displaystyle H$ is normal in $\displaystyle G$, otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...(the easiest example I can think of is $\displaystyle S_3$ - the elements of order $\displaystyle 2$ here do not generate normal subgroups. An example where the non-normal subgroups are of order not 2 would be in $\displaystyle S_4$ - 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

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Originally Posted by

**Swlabr** You are assuming that $\displaystyle H$ is normal in $\displaystyle G$, otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...

but $\displaystyle G/H $ is still a subgroup and it is order divides G order

and $\displaystyle |G/H| = |G|/|H| $ which is still even since H order is not even

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

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**Amer** but $\displaystyle G/H $ is still a subgroup and it is order divides G order

No. G/H denotes the SET of cosets $\displaystyle \{gH; g\in G\}$. They form a group IF AND ONLY IF H is normal in G.

For example, let $\displaystyle H=\{1, (1, 2)\}\leq S_3$. Then your cosets are $\displaystyle 1H$, $\displaystyle (2, 3)H$ and $\displaystyle (1, 2, 3)H$. They do NOT form a group - can you work out why not?

Also, $\displaystyle G/H$ doesn't form a subgroup of your group. Instead, it forms what is called a quotient group. For example, if you take the group of the integers under addition, then quotient out the subgroup $\displaystyle 2\mathbb{Z}=\{\ldots, -2, 0, 2, 4, \ldots,\}$ 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.

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

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Originally Posted by

**Swlabr** No. G/H denotes the SET of cosets $\displaystyle \{gH; g\in G\}$. They form a group IF AND ONLY IF H is normal in G.

For example, let $\displaystyle H=\{1, (1, 2)\}\leq S_3$. Then your cosets are $\displaystyle 1H$, $\displaystyle (2, 3)H$ and $\displaystyle (1, 2, 3)H$. 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

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**Swlabr** , if you take the group of the integers under addition, then quotient out the subgroup $\displaystyle 2\mathbb{Z}=\{\ldots, -2, 0, 2, 4, \ldots,\}$ 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

$\displaystyle \mathbb{Z} /2\mathbb{Z} = \{ 2\mathbb{Z} , 1+ 2\mathbb{Z} \} $

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

Thanks very much

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

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Originally Posted by

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

$\displaystyle \mathbb{Z} /2\mathbb{Z} = \{ 2\mathbb{Z} , 1+ 2\mathbb{Z} \} $

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 $\displaystyle G$ which has a normal subgroup $\displaystyle H$ such that $\displaystyle G/H$ is not isomorphic to any subgroup of $\displaystyle G$...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

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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 $\displaystyle G$ which has a normal subgroup $\displaystyle H$ such that $\displaystyle G/H$ is not isomorphic to any subgroup of $\displaystyle G$...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

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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 $\displaystyle G$ which has a normal subgroup $\displaystyle H$ such that $\displaystyle G/H$ is not isomorphic to any subgroup of $\displaystyle G$...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

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**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

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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

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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

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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 $\displaystyle h.n = hnh^{-1}$. 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:

$\displaystyle G/N = NH/N \cong H/(N\cap H) = H$, 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:

$\displaystyle \varphi:H \to \text{Aut}(N)$, 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 $\displaystyle |N|*|H|$ elements, so we take the cartesian product $\displaystyle N \times H$ 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:

$\displaystyle (n_1,h_1)(n_2,h_2) = (n_1\varphi_{h_1}(n_1),h_1h_2)$

where $\displaystyle \varphi_{h_1}$ is the automorphism of N induced by $\displaystyle h_1$, that is, the image $\displaystyle \varphi(h_1)$.

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:

$\displaystyle (n_1,h_1)[(n_2,h_2)(n_3,h_3)] = (n_1,h_1)(n_2\varphi_{h_2}(n_3),h_2h_3)$

$\displaystyle = (n_1\varphi_{h_1}(n_2\varphi_{h_2}(n_3)),h_1(h_2h_ 3)) = (n_1(\varphi_{h_1}(n_2)\varphi_{h_1}(\varphi_{h_2} (n_3)),(h_1h_2)h_3)$

$\displaystyle = ((n_1\varphi_{h_1}(n_2))(\varphi_{h_1h_2}(n_3)),(h _1h_2)h_3)$

$\displaystyle = (n_1\varphi_{h_1}(n_2),h_1h_2)(n_3,h_3) = [(n_1,h_1)(n_2,h_2)](n_3,h_3)$ (whew!)

well, a group needs an identity, so we'd better have one. the obvious candidate is $\displaystyle (e_N,e_H)$. so let's check:

$\displaystyle (n,h)(e_N,e_H) = (n\varphi_{e_H}(e_N),he_H)$

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

$\displaystyle = (ne_N,h) = (n,h)$

so we at least have a monoid. do we have inverses? given (n,h) consider the element $\displaystyle (\varphi_{h^{-1}}(n^{-1}),h^{-1})$. again, we compute:

$\displaystyle (n,h)(\varphi_{h^{-1}}(n^{-1}),h^-1) = (n\varphi_h(\varphi_{h^{-1}}(n^{-1})),hh^{-1})$

$\displaystyle =(n\varphi_{hh^{-1}}(n^{-1}),e_H) = (n\varphi_{e_H}(n^{-1}),e_H)$

$\displaystyle = (nn^{-1},e_H) = (e_N,e_H)$, 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 $\displaystyle G = N \rtimes_{\varphi}H$, to emphasize the fact that we are using the homomorphism $\displaystyle \varphi$). so we define:

$\displaystyle H' = \{(e_N,h): h \in H\}, N' = \{(n,e_H): n \in N\}$.

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

$\displaystyle (n,h)(n',e_H)(\varphi_{h^{-1}}(n^{-1}),h^{-1}) = (n\varphi_h(n'),h)(\varphi_{h^{-1}}(n^{-1}),h^{-1})$

$\displaystyle = ((n\varphi_h(n'))(\varphi_h(\varphi_{h^{-1}}(n^{-1})),e_H)$

$\displaystyle =(n\varphi_h(n')n^{-1},e_H)$, which is clearly in N'.

it's also clear that $\displaystyle N'\cap H' = \{(e_N,e_H)\}$, and by construction, G = N'H'.

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

but what does $\displaystyle hnh^{-1}$ mean in this context? we have to actually use H' and N', so we have:

$\displaystyle (e_N,h)(n,e_H)(\varphi_{h^{-1}}(e_N),h^{-1}) = (\varphi_h(n),h)(\varphi_{h^{-1}}(e_N),h^{-1})$

$\displaystyle =(\varphi_h(n),h)(e_N,h^{-1}) = ((\varphi_h(n))(\varphi_h(e_N)),e_H)$

$\displaystyle =(\varphi_h(n),e_H)$.

so we see that conjugating by $\displaystyle (e_N,h)$ really does give us the action:

$\displaystyle (n,e_H) \to (\varphi_h(n),e_H)$.

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 $\displaystyle \varphi: H \to \text{Aut}(N)$, 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:

$\displaystyle (n_1,h_1)(n_2,h_2) = (n_1n_2,h_1h_2)$

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

suppose that we have the cyclic group $\displaystyle \mathbb{Z}_n$, and the cyclic group $\displaystyle \mathbb{Z}_2$. besides the trivial homomorphism:

$\displaystyle \varphi(k) = \text{id}_{\mathbb{Z}_n}$ for k = 0,1, we also have the following homomorphism:

$\displaystyle \varphi(0) = \text{id}_{\mathb{Z}_n}$

$\displaystyle \varphi(1) = \text{inv}$, where inv is the automorphism that sends $\displaystyle m \to -m$.

in the latter case $\displaystyle \mathbb{Z}_n \rtimes_{\varphi} \mathbb{Z}_2$ has the following multiplication:

$\displaystyle (a,b)(a',b') = (a+(-1)^ba',b+b')$ 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 $\displaystyle rs = sr^{-1}$, which is just our old friend $\displaystyle D_n$ 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