Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

Question

I want to show without using Sylow theorem that

Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$

Please tell me whether I'm right!

I'm skecptical as I didn't use that $p$ is prime.

Answer 1

We only need to consider the case that $p$ is odd. If there is no element of order $p$ in the group, then every non-identity element of the group has order $2$. But a group in which every non-identity element has order $2$ is Abelian: ($1 = abab,$ so $aba = b$ and $ba = ab).$ The group has more than one subgroup of order $2,$ so since it is Abelian, it has a subgroup of order $4,$ contrary to Lagrange's theorem.

Answer 2

First why that subgroup of order two is normal? every group of order 2p need not have a normal subgroup of order 2. Consider $S_3$. So in your proof S need not form a group in general under coset product.

Answer 3

Why the set of left cosets of $H$ forms a group? The question is a special case of Every group has a subgroup of prime order?

Answer 4

By cauchy's theorem for every prime dividing order of the group there exists an element of order p. proof of cauchy's theorem requires no sylow theorems. consider the subgroup generated by that element to get a subgroup of order n.

Answer 5

Write $G$ for the group. If $G$ is cyclic, let $g$ be a generator, and we have that $g^2$ has order $p$. Otherwise, note that $G$ is solvable by Burnside's theorem. The only simple solvable groups are cyclic of order $p$, so we may write $1\lhd N \lhd G$. If $N$ has order $p$, we're done. If $N$ has order $2$, then take a nontrivial element $xN$ in $G/N$. Then $o(xN)$ divides $o(x)$, but $G$ is not cyclic, so it follows that $o(x)=p$.

Answer 6

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 8 on p.75 in Herstein's book.

I will use APPLICATION1 (CAUCHY'S THEOREM FOR ABELIAN GROUPS) on p.61 in Herstein's book.
Herstein proved this theorem without using Sylow's theorem.

APPLICATION1 (CAUCHY'S THEOREM FOR ABELIAN GROUPS):
Suppose $G$ is a finite abelian group and $p\mid o(G)$, where $p$ is a prime number. Then there is an element $a\neq e\in G$ such that $a^{p}=e$.

My solution is here:

First, consider the case $p=2$.
If $G$ is cyclic, then $G=\{e,g,g^2,g^3\}$ for some $g\in G$.
Then, $H:=\{e, g^2\}$ is a subgroup of $G$ whose order is $2$.
Assume that $G$ is not cyclic.
Then, $o(g)\in\{1,2\}\text{ }$ for all $g\in G$.
Let $g\in G-\{e\}$.
Then, $H:=\{e, g\}$ is a subgroup of $G$ whose order is $2$.

Second, consider the case $p$ is an odd prime number.
If $G$ is cyclic, then $G=\{e,g,\dots,g^{2p-1}\}$ for some $g\in G$.
Then, $H:=\{e, g^2,\dots,(g^{2})^{p-1}\}$ is a subgroup of $G$ whose order is $p$.
Assume that $G$ is not cyclic.
Then, $o(g)\in\{1,2,p\}\text{ }$ for all $g\in G$.
If there exists an element $g$ of $G$ whose order is $p$, then $H:=\{e,g,\dots,g^{p-1}\}$ is a subgroup of $G$ whose order is $p$.
Assume that $G$ is not cyclic and $G$ doesn't have an element whose order is $p$.
Then, $o(g)\in\{1,2\}\text{ }$ for all $g\in G$.
Then, $g^2=e\text{ }$ for all $g\in G$.
Then, $g^{-1}=g\text{ }$ for all $g\in G$.
So, $gh=(gh)^{-1}=h^{-1}g^{-1}=hg\text{ }$ for all $g,h\in G$.
So, in this case, $G$ is abelian.
Then, by APPLICATION1 (CAUCHY'S THEOREM FOR ABELIAN GROUPS) on p.61 in Herstein's book, $G$ has an element $g$ of $G$ whose order is $p$.
This is a contradiction.
So, it is impossible to hold $o(g)\in\{1,2\}\text{ }$ for all $g\in G$.

Let $H$ be a subgroup of $G$ whose order is $p$.
Then, $i(H):=(G:H)=2$.
So, $H$ is a normal subgroup of $G$.