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I really don't know how to solve this problem. I just know that if $|a|=2k_1+1$ and $|b|=2k_2+1$, then,

$a^{2k_1+1}=e=a^0$ and $b^{2k_2+1}=e=b^0$. Also, if $|G|=n$, then, $2k_1+1,2k_2+1\equiv 0 \pmod n$.

Shaun
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Hopmaths
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3 Answers3

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You got it the other way round. It's $n \equiv 0 \pmod{2k_1 + 1}$ and $n \equiv 0 \pmod{2k_2 + 1}$.

Suppose $a^2b^2 = b^2a^2$. Then: \begin{align*} ab &= a^{2k_1+1}(ab)b^{2k_2+1} \\ &= a^{2k_1+2}b^{2k_2+2} \\ &= (a^2)^{k_1+1}(b^2)^{k_2+1} \\ &= (b^2)^{k_2+1}(a^2)^{k_1+1} \\ &= b^{2k_2+1}(ba)a^{2k_1+1} \\ &= ba \end{align*}

Clement Yung
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    What if $a$ and $b$ have even order? – Hopmaths Jun 08 '20 at 18:46
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    @Heriberto then we can't conclude anything. Just take any $S_n$, and transpositions have order $2$, so if $a$ and $b$ are two distinct transpositions, then $a^2$ and $b^2$ are both $e$ and hence commute. However, distinct transpositions do not commute. – Clement Yung Jun 08 '20 at 19:01
  • How did you know that $(a^2)^{k_1+1}(b^2)^{k_2+1}=(b^2)^{k_2+1}(a^2)^{k_1+1}$? – kman May 14 '23 at 20:13
  • @kman Start by proving the identity for $k_1 = k_2 = 2$, and convince yourself that you can repeat the same proof for higher values. – Clement Yung May 15 '23 at 00:22
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Hint: $a = (a^2)^m$ if $a$ has order ...

Robert Israel
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  1. If $x$ commutes with $y$, then x commutes with every power of $y$: by induction, we have that if $xy^k=y^kx$, then $$xy^{k+1}=(xy^k)y = (y^kx)y = y^k(xy) = y^k(yx) = y^{k+1}x.$$ And if $x$ commutes with $y^k$, then it commutes with $y^{-k}$: from $xy^k=y^kx$, multiplying by $y^{-k}$ on both sides we get $y^{-k}x = xy^{-k}$.

  2. If $a^2$ commutes with $b^2$, then it commutes with every power of $b^2$.

  3. If $b$ has odd order, then $b$ is a power of $b^2$.

Arturo Magidin
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