why group of order six are not correct if we suppose that each element in the group are its own inverse?

Question

anyone help me but simply, because i am not math. boy i am chem.boy

Answer 1

You need to prove that a group where every element is of order two, has 2^n elements for some n.

This is simple for albelian groups.
It is harder for non-albelian groups, but can be done by induction over the number of elements.

I am sure that there are other ways of doing this too.

Answer 2

Let a be any element of G other than the identity, e.
and suppose that a = a^-1. Then a² = e so every
element of G except the identity has order 2.
But 3 is a prime dividing 6, so by one of the
Sylow theorems G contains a subgroup
of order 3, which is cyclic and is therefore
generated by an element of order 3.
Contradiction.

Answer 3

It's possible to do an elementary proof by contradiction.

Let's say three elements of the group are 1, a, b.

ab is not equal to 1, a, or b (that's easy to check). So call it c.

Then ac = aab = b. Also, cb = abb = a.

So cba = aa = 1. So ba = c-inverse = c.

Finally, bc = bba = a. That covers, I think, the multiplication table for 1, a, b, and c.

Call the other two elements d and e.

What's ad? It's not 1, a, b (which equals ac), c (which equals ab), or d. So it must be e. But similarly bd must equal e. But you can't have ad = bd.

So there's your contradiction. QED