This is an easy answer but i dont get it.

**sfspitfire23** · October 29th 2009, 05:17 PM

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

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

Thanks!

**tonio** · October 29th 2009, 08:32 PM

Originally Posted by sfspitfire23

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

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

Thanks!

A subroup of...who??

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

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

Tonio

**sfspitfire23** · October 29th 2009, 08:56 PM

So there indeed will be 2 inverses?

**tonio** · October 29th 2009, 09:15 PM

Originally Posted by sfspitfire23

So there indeed will be 2 inverses?

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

Tonio

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