Question

How do you prove that a group `G` is abelian if and only if `a^2 b^2 = (ab)^2` for all `a, b in G` ?

Answer 1

If a group `G` is Abelian, then for any two elements `a` and `b`,

`a*b = b*a`

Now,

Define `c=a*b` which is also a member of `G` on virtue of closure property.

Then `c^2 = c*c`
`implies (a*b)^2=(a*b)*(a*b)`

By associative axiom,

`(a*b)^2 = a*(b*a)*b`

But, by property of commutativity (for an Abelian group),

`implies (a*b)^2=a*(a*b)*b`
`implies (a*b)^2=(a*a)*(b*b)`

Thus we get our final result,

`(a*b)^2 = a^2*b^2`

For two arbitrary elements `a` and `b` of an Abelian group `G`.

Answer 2

See explanation:

Explanation:

Since group multiplication is associative I will omit the allowed rearrangements of parentheses that indicate performing multiplications in different orders.

Suppose `G` is abelian.

Then for all `a, b in G`:

`ab=ba`

Then, using (associativity and) commutativity we find:

`(ab)^2 = acolor(blue)(ba)b = acolor(blue)(ab)b = a^2b^2`

Conversely, suppose that for all `a, b in G`:

`(ab)^2 = a^2b^2`

Then for all `a, b in G`:

`ba = a^(-1)color(blue)(abab)b^(-1) = a^(-1)color(blue)(a^2b^2)b^(-1) = ab`

i.e. `G` is abelian.