# Thread: This is an easy answer but i dont get it.

1. ## This is an easy answer but i dont get it.

OK, $\displaystyle t_{ab}=ax+b$. I have to prove this is a subgroup of S4. Now, Ive already proved closure by composition $\displaystyle t_{ab}=ax+b$ with $\displaystyle t_{cd}=cx+d$ to yield $\displaystyle (ac)x+(ad+b)$.

Now, the identities wrt composition are $\displaystyle a=1$ and $\displaystyle b=0$. What is the inverse!? How would you find the inverse wrt composition here?

Thanks!

2. Originally Posted by sfspitfire23
OK, $\displaystyle t_{ab}=ax+b$. I have to prove this is a subgroup of S4. Now, Ive already proved closure by composition $\displaystyle t_{ab}=ax+b$ with $\displaystyle t_{cd}=cx+d$ to yield $\displaystyle (ac)x+(ad+b)$.

Now, the identities wrt composition are $\displaystyle a=1$ and $\displaystyle b=0$. What is the inverse!? How would you find the inverse wrt composition here?

Thanks!
A subroup of...who??

Anyway, it must be that $\displaystyle T_{\alpha \beta}T_{ab}=x \Longrightarrow (a\alpha)x+(a\beta +b)=x$

Well, now just solve the above for $\displaystyle \alpha\,,\,\,\beta$

Tonio

3. So there indeed will be 2 inverses?

4. Originally Posted by sfspitfire23
So there indeed will be 2 inverses?

Of course not: this is a group. What gave you that idea?

Tonio