I came across two definitions of a maximal subgroup here. The "symbol-free definition" reads:
- A maximal subgroup of a group is a proper subgroup such that there is no other proper subgroup containing it.
However, the "definition with symbols" reads:
- A subgroup $H$ of a group $G$ is termed maximal if $H \lneq G$ and if $H \leq K \leq G$ for some subgroup $K$, then either $H = K$ or $H = G$.
I am trying to understand how these two definitions are equivalent. I am having trouble in developing the "definition with symbols" from the "symbol-free definition". My attempt at understanding them is as follows:
Fix that $H$ is a proper subgroup of $G$, that is $H \lneq G$. Now, let $K$ be an arbitrary proper subgroup of $G$, that is $K \lneq G$. Then, $H$ is a maximal subgroup of $G$ if $K$ does not properly contain $H$, that is, $H = K$. But, since $K$ is a proper subgroup of $G$ by assumption, I cannot say that $K = G$ is another possibility. How can the "definition with symbols" be recovered from the "symbol-free" definition? Why is the "definition with symbols" not assuming from the get-go that $K$ is some proper subgroup of $G$, like the "symbol free definition" demands?
Also: I read on the same page a definition of maximal subgroups "in terms of group actions" which reads:
- In terms of group actions, a subgroup of a group is maximal if the natural group action on its coset space is primitive.
It sounds more like a theorem to me, personally. I would to read a proof for this "definition" "in terms of group actions", if it exists somewhere on Math.SE. I've tried looking it up many times now, but couldn't find a proof for this exact statement anywhere. I know there are many proofs for similar statements, but they are not what I am looking for. Thanks for any help!
The "definition with symbols" says that $H$ is a proper subgroup of $G$, and if $K$ is a subgroup containing $H$, it is either $H$ or $G$. Now, let's think about these two cases. If $K=G$, then it isn't proper. So another way s reading the "definition with symbols" is that any subgroup containing $H$ is either $H$, or isn't proper. That is, the only proper subgroup containing $H$ is $H$ itself. So there is no proper subgroup containing $H$. But this is exactly what the "definition without symbols" says.
Going the other way, if there is no proper subgroup containing $H$, that means for any $K$ with $H \leq K < G$ we cannot have $H < K$. This leaves only $K = H$ as an option.
I hope this helps ^_^