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.
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Title: For any divisor $k$ of $n$, right? – Dietrich Burde Apr 27 '19 at 18:40
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@ Dietrich Burde , I meant "for each", i will edit it. – Wiliam Mikayla Apr 27 '19 at 18:46
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As Dietrich Burde said, you mean "for each divisor $k$ of $n$". At the moment, you have not said what $k$ is. – Derek Holt Apr 27 '19 at 18:47
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4This is a variation of this problem. Putting $k=1$ shows that there is a unique subgroup $N$ of order $2$, so $N \le Z(G)$. Now you apply the linked problem to $G/N$ to deduce that $G/N$ is cyclic, and then deduce that $G$ is cyclic. – Derek Holt Apr 27 '19 at 18:49
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I will check that after i finish my job , thank you very much @Derek Holt. i will try write answer to my own question by my self here and ask you to chechk my solution if that is possible.Again thanks for your time. – Wiliam Mikayla Apr 27 '19 at 18:53
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Hi @Derek Holt , i have wrote my solution can you check my solution?thanks. – Wiliam Mikayla Apr 28 '19 at 19:24
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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.
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I think the final remark "it is easy to see that $G = \langle gn \rangle$ needs more explanation. Why is that easy to see? – Derek Holt Apr 28 '19 at 21:19
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Your final remark "we can write $G = \langle gn \rangle$" does not follow. It doesn't really make sense, because $g$ and $gn$ are both coset representatives of the coset $gN$, so whatever is true must remain true if you replace $g$ by $gn$. To prove this you need to consider the cases $n$ odd and $n$ even separately. – Derek Holt Apr 29 '19 at 07:16
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Well i am get confused i wrote $G= { e,g ,g^2, \cdots , g^{n-1},gn,g^2n,\cdots,g^{n-1}n }$ , is not it true ? If this is true then for $(gn)^k$ with $k \le ord(g)$ we can write $gngn...gn$ , $k$ times and since $n$ in the center of group we can change order of elements and write $g^kn^k$ now since $ord(n)=2$ if $k$ is even we have $g^kn^k=g^k$ and if k is odd we have $g^kn^k$=$g^kn$ so $
$ generates all the elements in $G$ , sorry i cant understand where is my flaw could you explain more? – Wiliam Mikayla Apr 29 '19 at 10:24 -
Yes, if $k$ is even then $g^kn^k = g^k$. Why should that imply that $G$ is cyclic? Could it not be the direct product $\langle g \rangle \times \langle n \rangle$? with $g$ having order $n$? And when $k$ is odd how do you conclude that $\langle gn \rangle$ generates all elements of $G$. You have written down a sequence of true statements, but then at the end you just state the desired conclusion with no explanation. – Derek Holt Apr 29 '19 at 11:33
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@Derek Holt i find out if $n$ is odd the $ord(gn)=2n$ so the claim is true but for even $n$ i checked example , $n=2$ at this point $G= {e,n,g,gn }$ but non of the elements generates $G$ so am i miss something ? Thanks for your patience and time. – Wiliam Mikayla Apr 29 '19 at 13:22
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Your statement about $n$ odd is still not completely right. In that case we have either $G = \langle g \rangle$ or $G = \langle gn \rangle$, but not both. When $n$ is even, you have to use the hypothesis again with $k=n/2$ to complete the proof· – Derek Holt Apr 29 '19 at 14:56
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@Derek Holt, is the folloing statement flawed? , $G \over N $ is cyclic thus there exist $g \in G$ such $ord(g)=n$ and ${G \over N} =
$ since we know that $G={ e,n,g,g^2,g^2n,\cdots,g^{n-1},g^{n-1}n }$ , if $n$ is odd then we claim $G= – Wiliam Mikayla Apr 29 '19 at 16:30$ to show this it is suffices to show $ord(gn)=2n$, we know $gn=ng$ and $gcd(ord(g),ord(n))=1$ and their order is finite so we can deduce that $ord(gn)=ord(g)ord(n)=2n$, now i think $n$ can not be even , suppose $n$ is even thus $n/2 | n$ and equation $x^n=e$ has at most $n$ solution but $(gn)^n=e$ and $(g^i)^n=e,1 \le i \le n$ which more than $n$ -
Yes it is flawed. It is not true that there must exist $g \in G$ of order $n$ with $G/N = \langle gN \rangle$. Also, what do you mean by saying that $n$ cannot be even? Clearly there exists a cyclic group of order $2n$ for any $n$. – Derek Holt Apr 29 '19 at 20:41
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@Derek Holt i am so sorry , it is only past one month since i start study algebra, excuse my silly mistakes, so i thinked more since we only know ${G \over N }$ is cyclic we only allowed to say there exist $gN \in {G \over N}, \ g \in G$ such that ${G \over N } =
$ , we know $n=Ord(gN) \ | \ Ord(g)$ and $G={ e,n,g,g^2,g^2n,\cdots,g^{n-1},g^{n-1}n }$ , now since $Ord(g) \ | \ 2n$ if $2n=Ord(g)$ it is clear $G= – Wiliam Mikayla Apr 29 '19 at 22:34$ and if $Ord(g)=n$ , since $Ord(n)=2$ , if $n$ is odd we obtain $Ord(gn)=2n$ thus $G= $ , is it corret for odd case? -
for the case $n$ even , since $n/2 \ | \ n$ using hypothesis the equation , $x^{2*(n/2)}=e$ has at most $n$ solutions in $G$ , we that for every $1 \le k \le n$ , that $(g^i)^n=e$ , but $(gn)^n=e$ which more than $n$ solution in $G$ thus $Ord(g) \neq n$ , this indicates that $Ord(g)=2n$ and thus $G=
$. am i right ? – Wiliam Mikayla Apr 29 '19 at 23:56 -
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@Derek Holt , thank you very much for your help , patience , and time. – Wiliam Mikayla Apr 30 '19 at 18:53