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Let $G$ be a group, and $a$, $b \in G$

  • $(bab^{-1})^{n} = ba^{n}b^{-1}$, for every positive integer $n$ \begin{align*} \text{Let P(n) be the statement: } (bab^{-1})^{n} &= ba^{n}b^{-1} \newline \text{Show the base case P(1) : } bab^{-1} &= bab^{-1} \newline \text{Assume P(k) is true : } (bab^{-1})^{k} &= ba^{k}b^{-1} \newline \text{Now I need to show P(k+1) is true by multiplying P(k) by } bab^{-1} \newline (bab^{-1})^{k}(bab^{-1}) &= ba^{k}b^{-1}(bab^{-1}) \newline &= ba^{k}(b^{-1}b)ab^{-1} \newline &= ba^{k}ab^{-1} \newline &= ba^{k+1}b^{-1} \newline \text{Which is the statement P(k+1)} \end{align*}

  • If $a^{-1}$ has a cube root, so does $a$.

If $a^{-1}$ does have a cube root, then there is an element $x$ in $G$ such that $a^{-1} = x^{3}$. \begin{align*} a^{-1} &= x^{3} \newline a^{-1}a &= x^{3}a \newline e &= x^{3}a \newline (x^{-1})^{3} &= (x^{-1})^{3}x^{3}a \newline (x^{-1})^{3} &= a \end{align*}

Comments on the correctness or ways to improve either proof would be appreciated :)

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As far as I can tell, those both look fine. – yunone Jul 28 '11 at 2:26
@Jon: They are all fine; but you can do the second one simpler by just taking inverses on both sides: if $a^{-1}=x^3$, then $a = (a^{-1})^{-1} = (x^3)^{-1} = (x^{-1})^3$. – Arturo Magidin Jul 28 '11 at 2:33
up vote 1 down vote accepted

Both of the proofs look good to me.

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HINT $\ \ \ $ Both answers follow from $\rm\ f(x^n) = f(x)^n\ $ for an (anti-) multiplicative map $\rm\:f\:.$

Slightly simpler and more general is to note that the map $\rm\ f(x) = b\:x\:b^{-1}\ $ is multiplicative, i.e. $\rm\:f(xy) = f(x)\:f(y)\:,\:$ so, by induction $\rm\ f(x^n) = f(x)^n\:.$

Similarly for the second problem $\rm\: f(x) = x^{-1}\: $ satisfies $\rm\ f(xy) = f(y)\:f(x)\:,\: $ therefore upon applying $\rm\ f\ \:$ to $\rm\ a^{-1}\! =\: x^n\ $ we infer that $\rm\ a\: =\: f(x^n) = f(x)^n = (x^{-1})^n\ \ $ (for $\rm\:n=3\:$ in your case).

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