# Proving $K$ is a group

Now I have proved certain things are a group before, and I know that it requires:

1)Associativity

2)Inverse

3)Identity

But here I have such a strange thing that I wanted to clarify that I am doing it right.

If $a,b \in \mathbb{R}$ with $a\ne0$, let $L_{a,b}:\mathbb{R} \to \mathbb{R}$ be the function given by $L_{a,b}(x) = ax+b$ for all $x$. Let

$$K = \{L_{a,b}|a,b\in \mathbb{R}\}$$

Prove that $K$ forms a group under composition of functions.

So normally you want to test some operation binary operation like $x+y = x+y+2xy$ or something, and see if it meets the above $3$ requirements. But here I don't know if I want to look at associativity in which of the following ways:

1) $((ax+b) + (cy+d))+(ez+f) = (ax+b) + ((cy+d)+(ez+f))$

2) $((ax+b) + (cx+d))+(ex+f) = (ax+b) + ((cx+d)+(ex+f))$

3) $((ax+b) + (ay+b))+(az+b) = (ax+b) + ((ax+b)+(ax+b))$

4) $((ax+b) + (ax+b))+(ax+b) = (ax+b) + ((ax+b)+(ax+b))$

Which one am I meant to be comparing?

How do I get the Inverse and Identity our? Does $L_{a,b}$ have some $x$ such that $ax+b=ax+b$, so just $x=x$ for identity, and some $ax+b=x$?

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Function composition is always associative. – lhf Jun 25 '14 at 14:58

We need to remember we the group operation is function composition, and not function addition.

So to establish associativity, we need to establish that $$(L_{a,b}\circ L_{c,d})\circ L_{e,f}=L_{a,b}\circ (L_{c,d}\circ L_{e,f})$$

Associativity

\begin{align} (L_{a,b}\circ L_{c,d})\circ L_{e,f} & = (L_{a, b}(cx + d))\circ L_{e, f}\\ & = [a(cx + d) + b] \circ L_{e, f}\\ & =[acx + ad+b]\circ L_{e, f}\\ & = L_{ac, ad+b}(ex + f) \\ &=ac(ex+f) + (ad+b) \\ &= a(cex+cf +d) + b\\ & =L_{a, b}\circ (cex + cf + d)\\ & = L_{a, b} \circ [c(ex+f) + d ]\\ &=L_{a, b} \circ(L_{c, d}( ex + f))\\ & = L_{a, b}\circ(L_{c, d} \circ L_{e, f}) \end{align}

Identity

$$L_{\text{id}}(x) = x = 1\cdot x + 0 = L_{1, 0}$$

Inverses

Let $L_{a, b}(x) = y =ax + b$ be any element in the group. Solve for $y$: $$y-b = ax \iff \frac{y-b}{a} = x$$

So $$L^{-1}_{a, b}(x) = \frac{x-b}{a} = \frac 1a\cdot x + \left(\frac{-b}a\right) = L_{\frac 1a, \frac{-b}a}$$ You can confirm that $L_{\frac 1a, \frac{-b}{a}}(x) = L^{-1}_{a, b}(x)$ by composing the two functions and obtaining the identity function: $L_{1, 0}$.

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None of them. Rememeber the group operation is composition of functions, not pointwise addition.

You need to check $(L_{a,b}\circ L_{c,d})\circ L_{e,f}=L_{a,b}\circ (L_{c,d}\circ L_{e,f})$ as functions of $x$.

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Does $x$ vary?${}$ – Incurrence Jun 24 '14 at 5:52
@user142198 Two functions $f$ and $g$ are equal iff $f(x)=g(x)$ for all $x$. – blue Jun 24 '14 at 5:53
@blue The "of $x$" is misleading, $L_{a,b}$ is a function of $x$ as well as a function of $t$, $z$ or whatever you want to name your variable. So "as functions" is enough. – Pece Jun 24 '14 at 6:00
@Pece Would I be right in saying that: Identity: $L_{c,d} = x$ Inverse: $L_{c,d} = \frac{x-b}a$ here? – Exam in 5 days Jun 24 '14 at 7:28
@Exam: It's strange to write $L_{c,d}$ for something that does not depend on $c$ or $d$… – Eric Stucky Jun 24 '14 at 9:06

Your question essentially boils down to: "What do elements of $K$ look like?" The answer is obvious from the definition of $K$ and $L_{a,b}$. Each element in $K$ is a degree one polynomial with arbitrary coefficients and constants in $\mathbb R$. So, we see that 2) is the claim you wish to prove. Also, note that the operation defined on $K$ is function composition and not function addition.

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+1 for "function composition", -1 for "the answer is obvious". – Eric Stucky Jun 24 '14 at 6:06