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By the center of a group $G$ we mean the set of all elements of $G$ which commute with every element of $G$, that is, $C = \{ a \in G: ax = xa \text{ for every } x \in G \}$.

We want to show that $C$ is a subgroup:

(i). Let $m, n \in C$, then $mx = xm$ and $nx = xn$ for every $x \in G$. Show that $(mn)x = x(mn)$. $mx = xm \rightarrow x = m^{-1}xm$ and $nx = xn \rightarrow x = n^{-1}xn$. Then substitute: $mnx = (mn)x = (mn)(n^{-1}xn) = mxn = m(m^{-1}xm)n = x(mn) = xmn$.

(ii). Let $m \in C$, then $mx = xm$. Show that $m^{-1}x = xm^{-1}$. From $mx = xm$ we can conclude that $x = mxm^{-1}$ and $x = m^{-1}xm$. So I need to show that $m^{-1}x = xm^{-1}$. We can substitute as follows: $m^{-1}x = m^{-1}(mxm^{-1}) = xm^{-1}$.

After solving the previous problem, I am having trouble making any progress on the following problem:

Let $C' = \{ a \in G: (ax)^{2} = (xa)^{2} \text{ for every } x \in G \}$. Prove that $C'$ is a subgroup of $G$.

(i). Let $m, n \in C'$, then $(mx)^{2} = (xm)^{2}$ and $(nx)^{2} = (xn)^{2}$. Show that $(mnx)^{2} = (xmn)^{2}$.

(ii). Let $m \in C'$, show that $m^{-1} \in C'$.

I've tried used a similar strategy to the center problem by solving for $x$ and multiplying $(mx)^{2}, (xm)^{2}, (nx)^{2}, (xn)^{2}$ in different ways. But I haven't made any progress. Could I get a hint for part (i)?

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Just to remind: to show that a subset of a group is a subgroup it is not enough to show that the subset is closed under the group and inverse operations. In addition one has to verify that the subset is nonempty. Usually this is done by showing that the neutral element is in the subset. – LostInMath Aug 16 '11 at 17:05
up vote 5 down vote accepted

I think you would find it less confusing if you weren't using $x$ for two different things.

In (i) you know that $(mx)^2 = (xm)^2$ for all $x\in G$ and that $(nx)^2 = (xn)^2$ for all $x\in G$. You want to show that $mn\in C'$. So, let $y\in G$, and we want to show that $\Bigl((mn)y\Bigr)^2 = \Bigl(y(mn)\Bigr)^2$. Well, $(mn)y = m(ny)$; setting $x$ equal to $ny$ we have $$\Bigl( (mn)y\Bigr)^2 = \Bigl( m(ny)\Bigr)^2 = \Bigl(mx\Bigr)^2 = \Bigl(xm\Bigr)^2 = \Bigl((ny)m\Bigr)^2.$$ Now notice that $(ny)m= n(ym)$ and do something similar.

Likewise, for (ii), you know that $(mx)^2 = (xm)^2$ for all $x$. Take $y\in G$, and look at $(m^{-1}y)^2$. Note that $$(m^{-1}y)^2 = \Bigl((y^{-1}m)^{-1}\Bigr)^2 = \Bigl( (y^{-1}m)^2\Bigr)^{-1}.$$ Now set $x=y^{-1}$.

P.S. And don't forget to check the sets are not empty!

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Thank you, your advice helped me complete the problem. To me this problem makes use of clever substitutions, and I doubt I would have been able to solve the problem myself. My soft-questions are: At what point should I ask on MSE for a hint/help? How much time should I be spending on these types of problems? And should I buy books that help me develop my skills in problem-solving (ie. olympiad books)? – Student Aug 16 '11 at 20:12
@Jon: Olympiad books are probably not the best way to develop the run-of-the-mill skills that are useful for standard mathematics: many of the problems turn instead on clever ideas or slick tricks rather than deep understanding. A solid textbook with lots of exercises is usually a better idea; for group theory, there's the relevant sections of Herstein's "Topics in Algebra" for a traditional approach, Rotman's "Introduction to the Theory of Groups" for a more modern one. As to how much time, that depends. So long as you show an honest effort and don't give up quickly, I don't think people mind – Arturo Magidin Aug 16 '11 at 20:16

For (i), write $$ (mnx)^2 = m(nx)m(nx). $$ Since for each $y \in G$ we have $mymy = ymym$, what can you do to the right-hand side of this? Do that, then do it again.

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Probably too late, for (ii)

$(gc^{-1})^2 = (egc^{-1})^2 = ((c^{-1}c)gc^{-1})^2 = (c^{-1}(cg)c^{-1})^2 = (c^{-1}(gc)c^{-1})^2 = (c^{-1}g(cc^{-1}))^2 = (c^{-1}ge)^2 = (c^{-1}g)^2$

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