-2
$\begingroup$

Ok so I know that generally,

$(gh)^{n}$ $\neq$ $g^{n}h^{n}$

because n-times for $(gh)^{n}$ is: $ghghghghghg$.

because n-times for $g^{n} h^{n}$ is $gggggggg hhhhhhhh$. (My professor made mention though that for abelian this may be true? Why).

(My professor also made mention that for $n=2, 12$ and $13$ will work? Why). Is it because

$[(12)(13)]^{2}$ = $12^{2}13^{2}$ ? It doesn't follow this problem unless

$12*13*12*13 = 12*12*13*13$?

Is this correct?

  • 0
    "for abelian this may be true? Why" What makes Abelian groups ... Abelian? What can you do in an Abelian group that you cannot necessarily do in general? – 2017-01-16
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    Did I make the right leap? My professor just spoke this. Is my leap ok . – 2017-01-16

2 Answers 2

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For $g,h\in G$, if $gh=hg$, then $(gh)^n=g^nh^n$ for all $n\in \Bbb{Z}$.
You can try to verify this using induction.

While if $gh\neq hg$, the result may not hold.
Take $G=S_3$ as example. $$(12)(13)=(132)$$ Then $$(12)^2(13)^2=1$$ but $$[(12)(13)]^2=(132)^2=(123)$$ Hence $$(12)^2(13)^2\neq [(12)(13)]^2$$

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We can take the group $M(2,\mathbb Z)$:

Let be $A=(\array{1&0\\0&2})$ and $B=(\array{1&0\\1&0})$ then $AB=(\array{1&0\\2&0})$ and $(AB)^n=(\array{1&0\\2&0})$

while $B^n=(\array{1&0\\1&0})$ and $A^n=(\array{1&0\\0&2^n})$ then $B^nA^n=(\array{1&0\\1&0})$.

We note that $(AB)^n\ne B^nA^n$