Say $1 = g^2$ is a Group, so $p\cdot p = 1,$ prove that that the group is commutative?

Question

I know a good proof, but i want to proof it in my way, im asking if its right. Say $1 = g^2$ is a Group, so $p\cdot p = 1$ with $p$ element of $P,$ prove that that the group is commutative?

• $1 = e,$ the neutral element.
• $1.$ so we know for every group exist associativity.
• $e = g\cdot g,$ from that we can conclude that every is the inverse element of itself.

i proof that $(g\cdot a) = (a\cdot g)$

$g\cdot a$ is also inverse to itself, so $(e\cdot g)\cdot(e\cdot g) = e.$ now the rest.

$$(g\cdot a)^2 \cdot (a\cdot g)^2 = e $$ $$(g\cdot a) \cdot(g\cdot a)\cdot (a\cdot g)\cdot(a\cdot g)=e$$

because of associativity $$(g\cdot a)\cdot g \cdot(a\cdot a)\cdot g)\cdot(a\cdot g)=e$$ $$g\cdot(a\cdot g)\cdot (g\cdot a)\cdot g=e$$ because of associativity i can also start with $(a\cdot g)$ and $(g\cdot a)$ and for that its only true if they are the inverse off another thus the same.

Answer 1

To be honest, I cannot make sense of your proof, although I suspect that is due to poor presentation. Let me try to rewrite what your proof is (if I guess your intentions correctly).

We need to show that for all $a,g\in G$, we have $ag=ga$. First, $(ag)^2=e$ and $(ga)^2=e$ implies that $(ga)^2(ag)^2=gagaagag=e$. Since $aa=e$, this means that $$gag(aa)gag=gaggag=g(ag)(ga)g=e.$$ Therefore, multiplying by $g$ on both sides, we get $g^2(ag)(ga)g^2=g^2$. But since $g^2=e$, this means $(ag)(ga)=e$, so $ag=(ag)^{-1}=ga$.

Note that my rewording of your proof is much clearer. In particular, the way you wrote it there are several logical connections which are not made clear, and so depending on the interpretation your proof could be considered incomplete or wrong. For example, once you have $g(ag)(ga)g=e$ you say that "I can also start with $ag$ and $ga$ and for that its only true if they are the inverse of each other thus the same", which doesn't really make sense to me---you have somehow jumped from the equation to $ag=(ga)^{-1}$ without justification. There are also quite a bit of unnecessary steps in between, e.g. I'm not sure why you needed to write $(eg)(eg)=e$, even though it is correct.

A much simpler proof along your lines would be simply noting that $(ag)(ga)=ag^2a=aea=a^2=e$. Within one step we got $ag=(ga)^{-1}$, and since every element is its own inverse this means $ag=ga$.