# Thread: Group Theory & Binary Operations

1. ## Group Theory & Binary Operations

Hey all,

Sorry if this is in the wrong spot, it just seemed to not fit elsewhere.

I'm having trouble with proving a binary operation, having just started the subject. I'm trying to do this question:-

"X = {(x,y):y=2x} subset $R^2$ and the map o: X x X --> $R$ is given by o: {(x,y), (a,b) = (x+a, y+b)}. Show that o is a binary operation and that (X,o) is a group."

Now, I know that to be a group, it needs to be associative, have an identity and an inverse, but I'm struggling with proving the binary operation part.

Can anyone provide some tips please?

2. i will just give u some hints. a general point of X looks like (a, 2a) (why ?)

take two such points .. act the binary operation upon them, see if this is again a point in X i.e, check whether it is again of the form (a, 2a).

3. Originally Posted by exphate
Hey all,

Sorry if this is in the wrong spot, it just seemed to not fit elsewhere.

I'm having trouble with proving a binary operation, having just started the subject. I'm trying to do this question:-

"X = {(x,y):y=2x} subset $R^2$ and the map o: X x X --> $R$ is given by o: {(x,y), (a,b) = (x+a, y+b)}. Show that o is a binary operation and that (X,o) is a group."

Now, I know that to be a group, it needs to be associative, have an identity and an inverse, but I'm struggling with proving the binary operation part.

Can anyone provide some tips please?
Is that really what the problem says? You say "X x X --> $R$" which is NOT a binary operation but then say that it maps to (x+a, y+b) which is a member of $X= R^2$. I am goig to assume that "X x X--> $R$" is just a typo and you mean "X x X--> X".
To show that is a binary operation, just use the definition of "binary operation" which is very simple: it is an operation applied to two things: any map from A x A to A, for any set A, is, by definition, a binary operation.