# Without Lagrange's theorem: If $H$ is a subgroup of odd order of $S_n$, then it is a subgroup of $A_n$.

Prove the following without Lagrange's theorem:

If $H$ is a subgroup of odd order of $S_n$, then it is a subgroup of $A_n$.

So this proof is pretty trivial if you have Lagrange's theorem, but we haven't covered that or cosets yet (our course is going super slow), so I probably shouldn't use it. I am having difficulty just deleting knowledge of the theorem from my brain, and I can't just reconstruct the relevant parts of Lagrange's theorem because it needs cosets. How might I prove this?

For reference, this is question 5.24 in Gallian's Contemporary Abstract Algebra 8th ed.

• In other words, you want to prove: if $H$ is a permutation group which contains an odd permutation, then the order of $H$ is even. Hint: what can you say about the number of odd permutations in $H$ versus the number of even permutations? – bof Oct 8 '13 at 6:01
• If $\sigma$ is a permutation of odd order, should it be in $A_n$? – Beginner Oct 8 '13 at 6:18
• @MarshalKurosh, a permutation of odd parity applied an odd number of times gives a permutation of odd parity. Since the identity permutation has even parity, it follows that any permutation of odd parity has even order, whence every permutation of odd order is even and a member of $A_n$. – Peter Taylor Oct 8 '13 at 7:05
• @Peter: You are right! I was asking the question to "fhyve" (which is almost same question he asked); but I forgot to add his name in my question. sorry! – Beginner Oct 9 '13 at 5:18
• By the way, Is this problem solvable by using Lagrange's Theorem? The theorem does just say that the order of H divides the order of $A_n$, but not say that H is a subgroup of $A_n$...... – NNNN Oct 13 '13 at 16:35

If $H$ is a subgroup of $S_n$, and if $H$ contains an odd permutation, then exactly half the elements of $H$ are odd permutations. (This is a standard exercise; e.g., see this answer.) If $H$ has odd order, it's impossible for exactly half of its elements to be odd permutations; therefore $H$ contains no odd permutations at all, i.e., $H$ is a subgroup of $A_n$.

Assume H contains an odd permutation. Then half of its members are even and half ot its members are odd. $|H|= |E| + |O| = 2* |E|$, which is a contradiction . Hence H lies in An

Another Approch :
Consider $\phi:H\to${$\pm 1$}
H is odd order In case is $\phi$ take both 1, -1 for H
Implies $|\phi(H)|=2$
$e\in H$ $\phi(e)=1$ Now
By first isomorphism theorem
$\phi(H)\times Kernel(\phi)=H$ implies $2/H$ # contradication to H is odd order
Hence $\phi$ take only 1.
That implies $H\subset A_n$