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Denote by $f_{a,b}$ the map from $\mathbb R$ to $\mathbb R$ such that $f(x)=ax+b$ for all $x \in \mathbb R$, where $a,b \in \mathbb R$ with $a\neq0$.

Let $G=\{f_{a,b}|a,b\in \mathbb R,a \neq0\}$. So I have already shown that $f_{a,b}$ is a bijection for any $a,b \in \mathbb R $ with $a\neq 0$ and prove that it is a subgroup of the group Sym$_{\mathbb R}$ for all bijective maps from $\mathbb R$ to $\mathbb R$.

Now I want to find if $G$ is an abelian group or not. I am aware abelian is the same as commutative, i.e., $ab=ba$.

However I am confused as to how I would apply this to my group. This is a group under composition, or at least that's was I was required to use to show it is a subgroup. I assume the same still stands to show whether or not it is commutative. In any case I have tried two methods, both of which could be wrong.

I tried to show: $$(ax+b)(a'x'+b')=(a'x'+b')(ax+b),$$ but under composition it would be trying to show: $$(a'(ax+b)+b')=(a(a'x'+b')+b).$$

But they both don't seem right to me, in any case I'm not really sure what the right direction is. Any help/hints would be very appreciated.

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  • $\begingroup$ You're right: you must do composition, you it'll be less confusing with a $\;\circ\;$ between those parentheses... $\endgroup$ – DonAntonio Mar 1 '17 at 16:56
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A group $G$ is abelian iff $gh = hg$ for all $g,h \in G$. Now, what are elements in your group? They are functions. So let us take two of them $g = f_{a,b}$ and $h = f_{c,d}$. Now calculate $gh$ and $hg$:

$$ gh(x) = g(cx + d) = a(cx + d) + b = acx + ad + b $$

$$ hg(x) = h(ax + b) = c(ax + b) + d = cax + cb + d $$

Now two functions are equal iff their value is same on every $x$. Now can you say whether $gh = hg$?

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  • $\begingroup$ @zahbaz yes, fixed! thanks $\endgroup$ – Vishal Gupta Mar 1 '17 at 17:05
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Is $\;f_{a,b}\circ f_{c,d}=f_{c,d}\circ f_{a,b}\;$ ? Meaning:

$$a(cx+d)+b\stackrel?=c(ax+b)+d\iff acx+ad+b\stackrel?=acx +bc+d$$

Now you produce a counterexample to deduce $\;G\;$ is non-abelian.

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It is about the second equation you posted (i.e. composition). But there is only one $x$ (no $x'$). Multuiplying it out then gives $$a'ax+a'b+b'=aa'x+ab'+b$$ which is not true in general. So no, the group is not abelian.

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You have to take 2 elements in $G$, say, $g$=$f_{a,b} $ and $g'$=$f_{a',b'}$, now check the composition: $g○g'(x)= f_{a,b} (a'x+b') = aa'x+ab'+b$ $g'○g (x)= f_{a',b'} (ax+b) =a'ax+a'b+b'$ From here is easy to construct a counterexample and show it is not abelian.

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