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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.

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I fixed up the formatting for you. Please consult a guide on LaTeX math editing to make any future questions more readable. The Not So Short Guide to LaTeX2e is a good starting point –  kahen Feb 15 '11 at 19:23
    
For (2), as all elements of $S_n$ have finite order, $H_n$ is a subgroup of $S_n$ if and only if it is (non-empty and) closed under multiplication. Did you already try multiplying some elements of odd order to see if their product has odd order again? Just play around multiplying various $3$-cycles of $S_5$ as a start. –  j.p. Feb 16 '11 at 9:12

1 Answer 1

For (1), since every subgroup of a cyclic group is itself cyclic, it is enough to show that the subgroup $K_0$ of $S_n$ generated by $a$ and $b$ is not cyclic. You should do some computations with a small example now, to see that $K_0$ is not cyclic.

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