# Opposite groups - show it is a group

1. Aug 15, 2010

### roam

1. The problem statement, all variables and given/known data

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.

3. 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?

2. Aug 15, 2010

### praharmitra

Re: Groups

ya, thats it.

3. Aug 16, 2010

### roam

Re: Groups

Really? Even the explanation for the closure is correct?

4. Aug 16, 2010

### HallsofIvy

Re: Groups

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.