Opposite groups - show it is a group

[roam]

Homework Statement

Suppose (G, \circ) is a group. Define an operation \star on G by a \star b = b \circ a for all a,b \in G. Show that (G,\star) is a group.

The Attempt at a Solution

So, I have to show that (G,\star) satisfies the associativity, identity, inverse and closure conditions:

Associativity: Let a,b,c be elements in G.

(a \star b) \star c = (b \circ a) \star c

= c \circ (b \circ a) = (c \circ b) \circ a

= a \star (b \star c).

Identity: (a \star e) = (e \circ a) = (a \circ e)=(e \star a) = a.

Inverse: y is an inverse of a then (a \star y)= (y \circ a) = (y \star a) = e.

Are these correct so far??

And for the closure, do I just need to show that any element of (G,\star) is in G? I know that (a \star b)=(b \circ a) and (b \circ a) \in (G, \circ) so is that all I need to say?

 

[praharmitra]

ya, that's it.

 

[roam]

Really? Even the explanation for the closure is correct?

 

[HallsofIvy]

Yes, (a \star b)=(b \circ a) and, because G is a group, \circ is closed so that b\circ a\in G for all a,b so a\star b\in G for all a, b.