I did it as follows: let order of G is n.let a,b be two non identity elements of G.If G is commutative then (ab) is a cyclic group other than (a) and (b),which is a contradiction,hence one of a and b must be of order n.so G is cyclic.If G is non commutative then G can't be cyclic.so we are done. This question is already asked in M S E and various answers are also given,but not this.Have I done any mistake?
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3Well...for one thing, how do you know $G$ is abelian? For another, if $b=a^k$ for some $k$ then $
$ needn't be different from $$ or $$. (or, if $b=a^{-1}$, we might have $ – lulu Oct 16 '17 at 13:23=e$). -
got it.but if i do: let G is not cyclic.so there exists one b s.t. it can't be written as power of a. then
is diffenent from them. – Anjan Samanta Oct 16 '17 at 13:33 -
"If G is non commutative then G can't be cyclic.so we are done." Why are you done then??? – amsmath Oct 16 '17 at 13:35
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So, maybe $G$ is not abelian. This is not as simple as you are making it out, though there are elementary proofs. See the responses to this question – lulu Oct 16 '17 at 13:36
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ok got it.if b is a inverse then there is a problem. – Anjan Samanta Oct 16 '17 at 13:40
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Let $O(G)=n$. Then $n$ has at most two distinct prime factors as three prime factor give three distinct elements of that order (Cauchy theorem) and hence three distinct subgroups. So $O(G)$ is either $p^k$ or $pq$ where $p,q$ are distinct primes. Case(i): If $O(G)=p$, though $G$ is cyclic but the condition failed. Case(ii): If $O(G)=p^2$ Since $G$ has two nontrivial subgroup, $G=H\cup K$ where $H,K$ are two distinct subgroup of order $p$. Equating number of elements we get $p^2=2p-1$ which cant be true. Case(iii): If $O(G)=p^3$. There exist subgroup $H$ of order $p$ and a subgroup $K$ of order $p^2$ by Sylow theorem on p-groups. They are unique and $K$ must have subgroup of order $p$, so $H\leq K$. This gives number of element of $G$ is atmost $p^2$, a contradiction. Case(iii) If $O(G)=p^k$($k\geq4$): Sylow theorems on p-group gives at least three subgroups of order $p,p^2,p^3$. Thus the condition failed. Case(iv) If $O(G)=pq$: There exist subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Those are unique and hence normal. Also $H\cap K=\{id\}$. So $G=HK$ is abelian and hence cyclic.
Nirmal Kotal
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