Let $G$ be a group. Let $a, b, c$ denote elements of $G$, and let $e$ be the neutral element of $G$. Prove the following.

Question

The problems are quoted. They are followed by proofs.

Please, check my work.

If $ab = e$, then $ba = e.$

By hypothesis, $a = b^{-1}$ and $b = a^{-1}.$ Then, $b^{-1}a^{-1} = e.$ So, $bb^{-1}a^{-1}a = bea.$ Thus, $e = ba.$

If $abc = e$, then $cab = e$ and $bca = e.$

Since $ab = e$, $ce = e$ and $c = e$. Then, $cab = ab = e$ and $bca = bea = ba = e.$

State a generalization of the two examples above.

If the product of two or more elements is equal to $e$, these elements commute.

If $xay = a^{-1}$, then $yax = a^{-1}.$

Let $xay = a^{-1}$. Then $xaya = e$. So, $yaxa = e$. Thus, $yax = a^{-1}.$

$a = a^{-1}$ iff $aa = e.$

$\to$

$(a)a = (a^{-1})a = e.$

$\leftarrow$

$(aa)a^{-1} = a = (e)a^{-1} = a^{-1}.$

$c = c^{-1} \to ab = c$ iff $abc = e.$

$\to$

$ab(c) = ab(c^{-1}) = cc^{-1} = e$.

$\leftarrow$

$(abc)c^{-1} = (e)c^{-1} \to ab = c^{-1}.$ Suppose $c = c^{-1}.$ Then, $ab = c.$

Let $a, b, c$ each be equal to its own inverse. If $ab = c$, then $bc = a$ and $ca = b$.

Let $ab = c$. Then, $bc = b(ab) = caaca= a.$ Also, $ca = abbc = ac = caacaab = b.$

If $abc$ is its own inverse, then $bca$ is its own inverse, and $cab$ is its own inverse.

Since $abcabc = e$, $cabcab = e$ and $bcabca = e.$

Let $a$ and $b$ each be equal to its own inverse. Then $ba$ is the inverse of $ab$.

Let $aa = e$ and $bb = e$, then $baab = e$.