# If X is a subset of G such that xy = yx for all x, y in X, then <X> is Abelian.

#### oblixps

##### Member
If $$X$$ is a subset of $$G$$ such that $$xy = yx$$ for all $$x,y\in X$$, then $$<X>$$ is Abelian.

I'm trying to understand the proof to the above statement in my book. This may be trivial but for some reason I am not seeing it.

It starts with saying $$X \subset C_{G}(X)$$ by the hypothesis and since $$C_{G}(X)$$ is a subgroup, we must have $$<X> \subset C_{G}(X)$$ and so $$X \subset C_{G}(<X>)$$. Then, just as above we have $$<X> \subset C_{G}(<X>)$$ and so <X> is abelian as desired.

i didn't understand how the book went from saying $$<X> \subset C_{G}(X)$$ and concluding that $$X \subset C_{G}(<X>)$$. it seems to me that $$X \subset C_{G}(<X>)$$ is clear from the hypothesis and i don't see why it was even needed to show that $$<X> \subset C_{G}(X)$$.

am i missing something here?

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#### Opalg

##### MHB Oldtimer
Staff member
Re: if X is a subset of G such that xy = yx for all x, y in X, then <X> is abelian.

i'm trying to understand the proof in my book. this may be trivial but for some reason i am not seeing it.

It starts with saying $$X \subset C_{G}(X)$$ by the hypothesis and since $$C_{G}(X)$$ is a subgroup, we must have $$<X> \subset C_{G}(X)$$ and so $$X \subset C_{G}(<X>)$$. Then, just as above we have $$<X> \subset C_{G}(<X>)$$ and so <X> is abelian as desired.

i didn't understand how the book went from saying $$<X> \subset C_{G}(X)$$ and concluding that $$X \subset C_{G}(<X>)$$. it seems to me that $$X \subset C_{G}(<X>)$$ is clear from the hypothesis and i don't see why it was even needed to show that $$<X> \subset C_{G}(X)$$.

am i missing something here?
The statement $Y\subset C_{G}(Z)$ means that $yz=zy$ for all $y$ in $Y$ and all $z$ in $Z$. That latter statement is completely symmetric in $Y$ and $Z$. It follows that $Y\subset C_{G}(Z)\;\Longleftrightarrow Z\subset C_{G}(Y).$

When you say "$X \subset C_{G}(<X>)$ is clear from the hypothesis", I think you are just saying that the equivalence $<X>\subset C_{G}(X)\;\Longleftrightarrow X\subset C_{G}(<X>)$ is obvious to you.

#### Deveno

##### Well-known member
MHB Math Scholar
it seems to me that all that is really being said here, is that if an element of G commutes with every element of a generating set S, for a subgroup H (that is H = <S>) it commutes with all of H.

so if every element of S commutes with every other, then H is abelian.

that is: it suffices to check if the set of generators is commutative, rather than the whole subgroup (perhaps a substantially savings of effort).

#### Opalg

##### MHB Oldtimer
Staff member
Maybe it will help to point out that there are two equivalent ways to define <S>, where S is a subset of some group.

1 (constructive definition). <S> consists of all products (of arbitrary length) formed from elements of S and their inverses.

2 (existential definition). <S> is the smallest subgroup containing S.

In this thread, you are told that each pair $x$, $y$, of elements of S commute with each other. If you are thinking in terms of definition 1., then to prove that <S> is abelian you only need to check that $x$ commutes with $y^{-1}$ (which is easy), and it's then kind of obvious that each pair of elements of <S> commute with each other.

But when giving formal proofs in group theory, it is usually more satisfactory to use definition 2., and that is what is done in the proof at the start of this thread. It is based on the fact that $C_G(S)$ is a subgroup which contains S, and therefore contains <S>.

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#### Deveno

##### Well-known member
MHB Math Scholar
i concur that in formal proofs, centralizers are a more elegant tool than explicit constructions (which can get messy, and hard to notate).

my comments were to show why this should be an "easy" proof, nothing is at all wrong with the proof or the method.