This question is from Kyoto university $2009 $ math exam .Although many of the question have long answer with explanation , the following question has not except for short answer. ($2009$-spring final exam question $3$ math $234$)

Let'say that m and n be real numbers and $m \neq0$ , the function $\gamma_{m,n}:R \rightarrow R$ is defined by $\gamma_{m,n}(x)=mx+n^2$. Let $\Gamma=\{\gamma_{m,n}:m \in R-\{1\},n \in R \}$ the sel of all functions of this type. Show that whether $\Gamma$ is a group under the composition of functions or not.

My work : I know that there are four condition to be a group such that clousure , associavity , identity and inverse. In these question , i said that it is a group but answer key says that it is not a group.

I think that answer key is wrong and it is a group. Is my answer true . If not ,can you say me why it is not a group

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    $\begingroup$ Please do not delete a question just after getting an answer. This is disrespectful to the person who took the time to answer your question, and to future readers who might have a related question, and for whom the posted answer may be helpful. $\endgroup$ – Xander Henderson Apr 16 at 19:14

Yes, you are correct. Indeed, you can represent the group of affine transformations of $\Bbb R$ as a subgroup of $GL(2,\Bbb R)$, namely $$\Gamma \cong \left\{\left[\begin{matrix} m & n \\ 0 & 1\end{matrix}\right]\right\}\subset GL(2,\Bbb R).$$

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    $\begingroup$ Take care, you haven't taken into account the fact that the upper right entry is positive: the set of matrices having the form $\left[\begin{matrix} m & \color{red}{n^2} \\ 0 & 1\end{matrix}\right]$ with $m,n \in \mathbb R$ isn't stable for multiplication. A couterexample with the initial context : composition of $x \to -x+1$ with $x \to -x+2$... $\endgroup$ – Jean Marie Apr 29 at 7:50
  • $\begingroup$ @JeanMarie The OP changed the question completely after I answered it (apparently, having deleted it first). $\endgroup$ – Ted Shifrin Apr 29 at 13:51
  • $\begingroup$ I thought something like that happened... $\endgroup$ – Jean Marie Apr 29 at 13:52
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    $\begingroup$ @JeanMarie: I have checked more carefully. The OP was editing as I typed my answer, then deleted the question because my answer was no longer applicable. What a mess. $\endgroup$ – Ted Shifrin Apr 29 at 15:25
  • $\begingroup$ It's difficult to imagine being at the place of such persons. But in some cases, maybe this one, such a reaction can be explained by a "face losing" issue. $\endgroup$ – Jean Marie Apr 29 at 15:34