If then, . On the other hand, Could you continue?
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?
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.
Strictly speaking, no. However, since the condition to be in the group is " ", which involves multiplcation, we would be pretty much assumed that the group operation is multiplication. Specifically, if and , then [tex](ab)^2= (a^2)(b^2)= (1)(1)= 1[/itex] while 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?
abab = aa bb = 1 a^2 = b^-2
b^-2 = 1
b^2 = 1
Is that what you wanted me to continue?
Thanks for your replies guys,
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