1. ## Comunitative Group

Hey,

I have a small question about groups,

If you have a comunitative 'group' H = <a in H : a2=1>,

Is that enough information to show that it is a group, without knowing the binary operation?

say b is also in H

then a*b=b*a

(a*b)*(b*a) = (a*a)*(b*b) = 1 (since its comunitative)

So that shows there is an identity, and each element is it's own inverse

It's also associative so everything is satisfied for H to be a group,

So only knowing these two properties of this group can show that it is indeed a group?

2. ## Re: Comunitative Group

If $\displaystyle x^2=1$ then, $\displaystyle x=x^{-1}$. On the other hand, $\displaystyle (ab)^2=1\Rightarrow abab=1\Rightarrow bab=a^{-1}\Rightarrow\ldots$ Could you continue?

3. ## Re: Comunitative Group

if H is commutative, then H is indeed a group if every element of H is of order 1 or 2. the commutativity is essential, since otherwise there is no guarantee that H is closed under multiplication. for example, the subset of S3 consisting of {e, (1 2), (1 3), (2 3)} is NOT a group.

as Fernando's post and mine taken together show, for such a subset H of some group G (that is, for all h in H, h^2 = e), closure under multiplication is equivalent to commutativity.

4. ## Re: Comunitative Group

Originally Posted by Angela11
Hey,

I have a small question about groups,

If you have a comunitative 'group' H = <a in H : a2=1>,

Is that enough information to show that it is a group, without knowing the binary operation?
Strictly speaking, no. However, since the condition to be in the group is "$\displaystyle a^2= 1$", which involves multiplcation, we would be pretty much assumed that the group operation is multiplication. Specifically, if $\displaystyle a^2= 1$ and $\displaystyle b^2= 1$, then [tex](ab)^2= (a^2)(b^2)= (1)(1)= 1[/itex] while $\displaystyle (a+ b)^2= a^2+ 2ab+ b^2= 2(1+ ab)$ which would not generally be equal to 1.

say b is also in H

then a*b=b*a

(a*b)*(b*a) = (a*a)*(b*b) = 1 (since its comunitative)

So that shows there is an identity, and each element is it's own inverse

It's also associative so everything is satisfied for H to be a group,

So only knowing these two properties of this group can show that it is indeed a group?

5. ## Re: Comunitative Group

abab = aa bb = 1 a^2 = b^-2

b^-2 = 1
b^2 = 1

Is that what you wanted me to continue?

I've been questioning the associativity a little more, I kind of just guessed it was associative,

Would this be sufficient to show associativity?

(a*b)*c=a*(b*c)

(a*c)*b= b*(a*c)

(a*c)*b= (a*c)*b

EDIT:

I'm sure the way my lecturer writes binary operations is a*b=ab (to save writing the star every time, so really ab is a*b), sorry I should of specified that, I thought it might have been the way everyone writes it