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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?

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    $\begingroup$ Looks good to me, very succinct. $\endgroup$ Jan 19, 2015 at 3:32
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    $\begingroup$ Yeah it is correct. $\endgroup$
    – corindo
    Jan 19, 2015 at 3:34
  • $\begingroup$ Okay, thanks for the verification $\endgroup$
    – St Vincent
    Jan 19, 2015 at 3:35
  • $\begingroup$ Yes, fine. Seems good. I like the idea you started. $\endgroup$ Jan 19, 2015 at 3:36

1 Answer 1

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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. $~~\blacksquare$

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