Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $ax+b=x$?

share|improve this question
    
Function composition is always associative. –  lhf Jun 25 at 14:58

3 Answers 3

up vote 3 down vote accepted

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: $L_{1, 0}$.

share|improve this answer

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 $x$.

share|improve this answer
    
Does $x$ vary?${}$ –  Committing to a challenge Jun 24 at 5:52
    
@user142198 Two functions $f$ and $g$ are equal iff $f(x)=g(x)$ for all $x$. –  blue Jun 24 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 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 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 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.

share|improve this answer
3  
+1 for "function composition", -1 for "the answer is obvious". –  Eric Stucky Jun 24 at 6:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.