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This is problem on Rotman's Exercise 3.35.

If $H<G$, then there exists a group $L$ and distinct homomorhpism $f,g:G \to L$ with $f|_{H} \neq g|_{H}$. (Hint. Let $L= S_{x}$, where $X$ denotes the family of all the left cosets of $H$ in $G$ together with an additional element $\infty$. If $a\in G$, define $f_{a} \in S_{x}$ by $f_{a} (\infty) = \infty$ and $f_{a}(bH) = abH$; define $g:G \to S_{x}$ by $g = \gamma \circ f$ where $\gamma : S_{x} \to S_{x}$ is conjugation by the transposition which interchanges $H$ and $\infty$)

As you can see, actually Hint is just answer. But I think the hint is false answer; For any $h \in H$, $f_{h}(H) = hH = H $. Hence when we represent $f_{h}$ as disjoint cycle representation, then it has 1-cylce (H) as its member. Also, clearly $(\infty)$ is also another member of $f_{h}$. Then, conjugation by $\gamma$ cannot affect $f_{h}$ since it just change 1-cycle member, so conjugation is disjoint with $f_{h}$ so it is just the same as $f_{h}$. Hence $f|_{H} = g|_{H}$. Therefore I think the hint is false.

Is there something I mistake or think wrong?

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  • $\begingroup$ The original exercise has a little omission: it must be $\;H\neq 1\;$ , of course. $\endgroup$ – DonAntonio Jan 23 '14 at 4:37
  • $\begingroup$ @DonAntonio Yes, but if we assume that $H$ is not trivial subgroup, I think for any nonidentity member $h \in H$, $hH = H$. Hence $f_{h}$ can't permute $H$. $\endgroup$ – user122655 Jan 23 '14 at 5:47
  • $\begingroup$ I think you will find that the exercise is to find $f,g$ with $f|_H = g|_H$ rather than $f|_H \ne g|_H$! And you are told that $H \ne G$. $\endgroup$ – Derek Holt Jan 23 '14 at 8:53
  • $\begingroup$ @DerekHolt I find it that the exercise clearly says find two distinct homomorphism with $f|_H \neq g|_H$. $\endgroup$ – user122655 Jan 23 '14 at 9:41
  • $\begingroup$ I have an older edition of Rotman, in which it is 3.49, and it definitely says $f|_H = g|_H$. It seems clear that your version is wrong and, as you noticed yourself, the hint answers my version. In any case, for $H \ne 1$, your version is easy, because you can take $L=G$, $f$ identity map, $g$ trivial map. $\endgroup$ – Derek Holt Jan 23 '14 at 10:28

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