Lemma on Permutations clarification

[PsychonautQQ]

Homework Statement

If k is moved by σ, then σk is also moved by σ

proof: otherwise σk is fixed by σ, that is σ(σk) = σk. But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis.


I am confused trying to understand this. I don't understand the part that says "But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis." I am not ignorant of what one-to-one means, but am having trouble putting together why that gives σk = k.

It seems logical to me you could have σ to a mapping that defines (1,2,3) → (4,5,6)
where k = {1,2,3} so σk(1) = (4) and σ(σk(1) = σ(4) = not in k

Homework Equations

The Attempt at a Solution

 

[pasmith]

Homework Statement

If k is moved by σ, then σk is also moved by σ

Please define your variables. My natural assumption is that $k \in \{1,\dots, n\}$ and $\sigma : \{1,\dots,n\} \to \{1,\dots,n\}$ is a permutation.

proof: otherwise σk is fixed by σ, that is σ(σk) = σk. But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis.

I am confused trying to understand this. I don't understand the part that says "But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis." I am not ignorant of what one-to-one means, but am having trouble putting together why that gives σk = k.

Let $\sigma(k) = j$. We have that $\sigma$ is one-to-one, so by definition for any $i \in \{1,\dots,n\}$, if $\sigma(i) = j$ then $i = k$.

If $\sigma(j) = j$ then, by the above, $j = k$.

 

[AlephZero]

It seems logical to me you could have σ to a mapping that defines (1,2,3) → (4,5,6)

That mapping is not a permutation. You also need to define what 4, 5, and 6 map to.

Another way to see why the proof works is to multiply by the inverse permutation. That gives σ^-1σ(σk) = σ^-1σk