0
$\begingroup$

Ok, so there is this problem that quite confuses me because the example seems too easy (at least thats what I think):

Find an example of a group homomorphism $\Phi: S_3 \rightarrow S_4$ that is injective.

I think the example is just $\Phi(\sigma)=\sigma$, this is the only example I can think about but I am not sure if it is correct. Can anyone tell me if this example is valid? If not what other examples could there be?

$\endgroup$
  • $\begingroup$ If $\sigma\in S_3$, then how can $\sigma$ be in $S_4$? $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 8:06
  • $\begingroup$ Technically $\sigma \in S_3$ is also in $S_4$ no? $\endgroup$ – Aurora Borealis Dec 9 '17 at 8:46
1
$\begingroup$

Your example is (almost) right. If you concretely take $S_n$ as the permutations of $\{1,2,...,n\}$ then any $\sigma \in S_3$ can be mapped to $g(\sigma) \in S_4$ by having $g(\sigma)(4)=4,$ while for $k=1,2,3$ put $g(\sigma)(k)=\sigma(k).$

$\endgroup$
  • $\begingroup$ Wait so I dont get it, is it correct to say it is $\Phi(\sigma)=\sigma$ is indeed an injective homomorphism, for all $\sigma \in S_3?$ $\endgroup$ – Aurora Borealis Dec 9 '17 at 8:47
  • $\begingroup$ @AuroraBorealis Not technically, since you want $\Phi(\sigma)$ to be in $S_4.$ And if $\sigma$ is in $S_3$ that's not in $S_4,$ unless in a loose sense in which you always map $4$ to itself. $\endgroup$ – coffeemath Dec 9 '17 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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