I was reading How to Think about Abstract Algebra by Lara Alcock, and encountered the following theorem and proof in the section on Lagrange's Theorem:
Claim: Let $K = \{e, (12)(34), (13)(24), (14)(23) \trianglelefteq S_4 \}$. Then for every $g \in S_4, g^6 \in K$.
Proof: Because K is normal, $S_4/K$ is a quotient group.
By Lagrange's Theorem, there are 24/4 = 6 cosets of $K$ in $S_4$.
So the quotient group $S_4/K$ has order 6.
Let $g \in S_4$ be arbitrary and note that $S_4/K$ has element $gK$.
Then, by the previous theorem [which states that, if group G has order $n$, then $g^n=e$ for all $g \in G$], $(gK)^6$ is equal to the identity in $S_4/K$.
But the identity in a quotient group is the subgroup, in this case $K$.
So $(gK)^6 = K = g^6K = K$.
So $g^6 = K$ for every $g \in G$.
I don't understand exactly what the highlighted sentence means or why it is, though. Can someone explain this to me?
A quotient group $S/K$ is a group where the elements are the cosets of $K$ in $S$. (Left or right cosets? Doesn't matter, quotient groups are only constructed when $K$ is normal, and thus right cosets are left cosets and vice versa, and we can just talk about cosets without specifying direction.) The identity element in $S/K$ is the coset of $K$ that is just $K$ itself.