Show that order of a Group is 2

Question

For a group $(G,*,I)$ if $a,b \in G$ such that $a * b$ has order 2. I need to prove that $b * a$ has order 2.

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My Work:-

$a * b$ has order 2 means $(a * b)^2 =I$ ,

so i need to prove is $(b * a)^2 =I$

$b * a * b * a =I$ $\qquad$ { expanding it }

$ a * b * a =b^{-1} * I$

$ a * b * a * b =b^{-1} * I * b$

$ a * b * a * b =b^{-1} * b$

$ a * b * a * b =I$

$ (a * b)^2 =I$

$ I =I$ $\qquad$ { given is $(a * b)^2 =I$ }

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Have I proved it in a correct way?

Answer 1

Sort of; it's kind of backward. You should start with what is known, and work toward your end goal. Otherwise, if you apply a non-invertible operation to both sides, you could end up with a specious proof.

$$ (a * b)^2 = I $$ $$ a * b * a * b = I $$ \begin{align} b * a * b * a * b & = b * I \\ & = b \end{align} \begin{align} b * a * b * a & = b * a * b * a * b * b^{-1} \\ & = b * b^{-1} \\ & = I \end{align}