Assume that $(G, \ast)$ is a group and that every element $a \in G$ satisfies $a \ast a = 1$. Show that $(G, \ast)$ is abelian.

Question

Assume that $(G, \ast)$ is a group and that every element $a \in G$ satisfies $a \ast a = 1$. Show that $(G, \ast)$ is abelian.

To prove $(G,\ast)$ is abelian, we must show that it is commutative. Let $a,b \in G$. Then $a \ast b \in G$ and $(a \ast b) \ast (a \ast b) = 1$. Taking both sides of the preceding equality by $\ast (b \ast a)$, we obtain $\begin{align} (a \ast b) \ast (a \ast b) \ast (b \ast a) &= 1 \ast (b \ast a) \\ (a \ast b) \ast (a \ast (b \ast b) \ast a)&=\\(a \ast b) \ast (a \ast a)&= \\a \ast b &= b \ast a \end{align}$

Is this correct?

Looks good to me, very succinct. – Travis Willse Jan 19 '15 at 03:32 — Travis Willse, Jan 19 '15 at 03:32
Okay, thanks for the verification – St Vincent Jan 19 '15 at 03:35 — St Vincent, Jan 19 '15 at 03:35
Yes, fine. Seems good. I like the idea you started. – Landon Carter Jan 19 '15 at 03:36 — Landon Carter, Jan 19 '15 at 03:36

Adriano · Accepted Answer · 2015-01-19T05:26:31.380

Looks great! Here's another proof that uses the fact that for any $g,h \in G$, we have that:

$$ g * g = 1 \iff g = g^{-1} \tag 1 $$ $$ (g * h)^{-1} = h^{-1} * g^{-1} \tag{2} $$

Given any $a, b \in G$, observe that: \begin{align*} a * b &= (a * b)^{-1} &\text{by (1), since } (a * b) * (a * b) = 1 \\ &= b^{-1} * a^{-1} &\text{by (2)} \\ &= b * a &\text{by (1), since } a * a = 1 = b * b \\ \end{align*} as desired. $~~\blacksquare$

Assume that $(G, \ast)$ is a group and that every element $a \in G$ satisfies $a \ast a = 1$. Show that $(G, \ast)$ is abelian.

1 Answers1