# A group G is abelian if

1. Feb 22, 2009

### hsong9

1. The problem statement, all variables and given/known data
show that a group G is abelian if g^2 = 1 for all g in G. Give an example showing that converse is false.

3. The attempt at a solution
Suppose g^2 = 1.
gg = 1,
(g^-1)gg = g^-1
g = g^-1 -- means self inverse. but I'm not sure how to show G is abelian..
and I don't know how I find counterexample.

2. Feb 22, 2009

### Daettil

Try taking 3 elements from G, say g, h, and gh. You should be able to show gh=hg pretty easily.

3. Feb 22, 2009

### Tobias Funke

Just to add one more thing to the above post, you've shown that g=g^(-1) for all g. In particular, gh=(gh)^(-1)=...?

The converse statement is "G is abelian implies g^2=1 for all g in G". You should have seen enough examples of groups by now to find an abelian group where this doesn't hold.

4. Feb 24, 2009

### hsong9

So.. Suppose that gg = 1, then ggh = hgg for all g,h in G, this clearly is that G is abelian.., right??
And counterexample...
G = {4,8,12,16}; multiplication in Z_(20)..
because g^2 not = 1 for all g in G, correct??
Also 1 means identity in this problem? or only number 1..?
Thanks

5. Feb 24, 2009

### VeeEight

Use the above poster's advice: gh=(gh)^(-1) by your discovery that each element is it's own inverse in this group.
Now compute (gh)^-1

For the counterexample perhaps you should consider Z_5

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