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$\begingroup$

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)?

  • 7
    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 verif$y$ that the subset is nonempty. Usually this is done by showing that the neutral element is in the subset.2011-08-16

3 Answers 3

1

Probably too late, for (ii)

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

5

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.

5

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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    @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 mind2011-08-16