groups again

[astronut24]

if a group of order 2p ( p prime) is abelian....then does it have exactly one element of order 2 ?? if a group is non abelian...i could figure out that there are p elements of order 2. but the abelian case is a bit confusing...
also..is it like...any group of order 2p has an element of order p?

if a group has orer p^a , a>=1 where p is prime....then i've gotta show that G has an element of order p.
can i say that any non-identity element in G can have order p or p^2 or p^3......or p^a. then if x in G is of the form x^(p^i) =e ....we can say, (x^(p^(i-1))^p= e and we've found an element x^(p^(i-1)) that is of order p....it seems too simple to be correct.

[mansi]

suppose G is abelian and has 2 elements of order 2, say a and b...then {e,a,b,ab} becomes a subgroup of G. but 4 does not divide 2p unless p=2. if p=2 then anyway {e,a} is a subgroup of G of order 2( p in this case).

[astronut24]

thanks for the help!
here's another question on the same lines: if G is a group where a^2=e for each a in G. show that order of G is 2^n for some n >=0. it's clear that the group is abelian and also clear that 2 divides O(G). how do you proceed further? by any chance, is it induction that we're supposed to use?

[matt grime]

The first answer also follows by the observation that any abelian group of order 2p for p an odd prime is cyclic.

The second follows from the fact that if p is a prime and p divides |G| then there is an element of order p in G.

[astronut24]

if G is a group such that a^2=e for each a in G. show that order of G is 2^n for some n>=0.
please help...
is the group of 3 non-singular upper triangular matrices a normal subgroup of
GL(3,R), the group of 3 cross 3 non singular matrices over R.

[matt grime]

I've answered the a^2=e for all a in G one.

As for the p^r one, all elements have order dividing p^r say the order of g is p^s, then p^{s-1] has order p.

[matt grime]

And why don't you just do the matrix one? it simple boils down to multiplying matrices together.

Hint: it's probably easier to find a counter example.

[matt grime]

Just to put here another useful result: if G is any group and all of its elements have order 2 then it is abelian. proof an exercise (hint what is (xy)^{-1})