1) Let $p$ be an odd prime, $n$ an integer such that $n \geq 3p$ and $a$ a $2p$-cycle in $S_n$.
Let $b$ be a $p$-cycle in $S_n$ and assume that $a$ and $b$ are disjoint. Suppose $K$ is a subgroup of $S_n$ that contains $a$ and $b$. Prove that $K$ is not cyclic.
(we may assume $a = (1\;\: 2\;\: 3\;\: \dotsb \;\:2p)$ and $b = (2p+1\;\: 2p+2\;\: \dotsb \;\:3p)$).
2) Let $H_n = \lbrace a \in S_n \;|\; |a|$ is odd$\rbrace$. For which values of $n\geq 2$ is $H_n$ a subgroup of $S_n$?
I have trouble starting number 1 because of its ambiguous definition of $H$. Let's start by a proof by contradiction approach and assume that $H$ is cyclic. Then do we define $H = \langle ab \rangle = \{e,ab,a^2b^2,a^3b^3,\dotsc\}$ and so forth? Or are they defined as $H = \langle a \rangle\langle b \rangle$? Does $H$ contain any other elements besides $a$ and $b$? If so how do we treat that?
For number 2 let's first note that every element in $H_n$ is in $A_n$ (proved in other part). Then $H_n$ is a "subset" of $A_n$. I suppose I can go about to prove this by saying every element of $H_n$ is in $A_n$ and get something out of that, but I can't quite piece it all together.
Thanks in advance.