if $|G|=2n$ and for each divisor of $n$, set $\{g \in G ; g^{2k}=e \}$ has at most $2k$ elements show that $G$ is cyclic. Hi guys i need a little help with this problem , I tried to use contradiction , set $G=\{g_1,\cdots,g_{2n} \}$ , I assumed by contrary that $G$ is not cyclic so there is not any $g \in G$ such that $Ord(g)=2n$, then I assumed $\max \{g_1,\cdots,g_{2n} \} = M$ , i defined $H_k= \{ g \in G ; g^{2k}=e \}$ , so $\exists \ 1 \le i \le 2n$ such $g_i^{M}=e$. But I stuck at this point and can not proceed any further. Is it possible to give me a hint? Thanks.
 A: since $2 \ | \ |G|=2n$ and 2 is a prime number using cauchy one can obtain that $G$ has a sub group with order $2$,suppose $N<G$ with $|N|=2$ by the problem hypothesis and setting $k=1$ we know that $G$ hasa t most $1$ sub group with order $2$ , so $N$ is a unique subgroup , suppose $N=\{e,n\}$ , $e$ is the identity element, usin the fact that , if a finite group $G$ has exactly one subgroup $N$ of a given order, then $N$ is a normal subgroup of $G$, so we can create quotient group $G \over N$ , now since $G$ is finite we know that $| {G \over N} |= n $ so suppose ${G \over N}=\{N,g_1N,\cdots,g_nN\}$, using this problem and knowing that equation $x^nN=N$ has at most $n$
solution in $G \over N$ we obtain $G \over N$ is cyclic , so set ${G \over N}=<gN>$ now since $|{G \over N}|=[G:N]=n$ and $G=N \cup g_1N \cup \cdots \cup g_nN$, by knowing that ${G \over N}= \{ N,gN,g^2N,\cdots ,g^{n-1}N \}$,we can write $G= \{e,n \} \cup \{ g,gn \} \cup \{g^2,g^2n \} \cup \cdots \cup \{g^{n-1},g^{n-1}n \} $ thus $G=\{ e,n,g,g^2,g^2n,\cdots,g^{n-1},g^{n-1}n \}$, now we know that $n=Ord(gN) \ | \ Ord(g)$ , if $2n=Ord(g)$  it is clear $G=<g>$ and if $Ord(g)=n$ since $Ord(n)=2$ , if $n$ is odd we obtain $Ord(gn)=2n$ thus $G=<gn>$, now if $n$ is even  again $n=Ord(gN) \ | \ Ord(g)$ and $Ord(g) \ | \ 2n$ so $Ord(g)=n$ or $Ord(g)=2n$ , now if $Ord(g)=n$ since $n/2 \ | \ n$ using hypothesis the equation , $x^{2 \times (n/2)}=e$ has at most $n$ solutions in $G$ , we know that for every $1 \le k \le n$ we have $(g^i)^n=e$ but $(gn)^n=e$ which is more than $n$ solution in $G$ thus $Ord(g) \neq n$ , this indicates that $Ord(g)=2n$ and thus $G=<g>$. this proofs the claim.
