0
$\begingroup$

Let $A$ be a finite set, and $B$ a subset of $A$. Let $G$ be the subset of $S_A$ consisting of all the permutations $f$ of $A$ such that $f(x)\in B$ for every $x\in B$. Prove that $G$ is a subgroup of $S_A$.

Since $A$ is a finite set we only have to prove that $G$ is closed under the operation. But I am unsure of how to go about showing that. I know that I have to show that $g_1\circ g_2\in G$ for $g_1,g_2\in G$.

$\endgroup$
  • $\begingroup$ If $x\in B$, then $g_2(x)\in B$, and hence $g_1(g_2(x)) \in B$ $\endgroup$ – Prahlad Vaidyanathan Oct 28 '13 at 18:01
  • $\begingroup$ Okay thanks for wording that. I knew it was along those lines but I couldn't figure out to word it correctly. :) $\endgroup$ – Student Oct 28 '13 at 18:04
0
$\begingroup$

If $x\in B$, then $g_2(x)\in B$, and hence $g_1(g_2(x))\in B$ – Prahlad Vaidyanathan Oct 28 at 18:01

To beef up this "answer", I'll add that the statement may fail for infinite sets. For example, if $A=\mathbb Z$ and $B$ is the set of positive integers, then $g(x)=x+1$ has the property $x\in B\implies g(x)\in B$ but its inverse does not.

| cite | improve this answer | |
$\endgroup$

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.