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$G$ is an abelian group of order $n$ with the property that $G$ has at most $d$ elements of order $d$, for any $d$ dividing $n$. Then $G$ is cyclic.

I am not getting any clue how to start. Please Help.

Tom Bombadil
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User8976
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  • Do you have the classification of finitely generated abelian groups? They are isomorphic to products of cyclic groups with prime powers as order. Now the product of two finite cyclic groups is cyclic iff they have coprime order. Can you make it from here? –  Aug 07 '16 at 13:22

2 Answers2

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First, decompose $G=C_1\times\cdots\times C_n$, via the Fundamental theorem of finitely generated abelian groups, where the $C_n$ are cyclic. Let's show that the orders of the $C_i$ are pairwise coprime. Let's start, say, with $C_1$ and $C_2$.

Suppose $d$ divides both $|C_1|$ and $|C_2|$. Then $C_1$ has an element $\alpha$ of order $d$, and $C_2$ also has an element $\beta$ of order $d$. Then the elements of the form $(\alpha,\beta^i)$ and $(\alpha^i,\beta)$, $i=0,\ldots,d-1$ have order $d$. Thus we've found $2d-1$ elements of order $d$, so the hypothesis implies $2d-1\leq d$ and thus $d=1$.

Thus, $|C_1|$ and $|C_2|$ are coprime, and therefore $D_1:=C_1\times C_2$ is cyclic (Product of two cyclic groups is cyclic iff their orders are co-prime).

But now we have $G=D_1\times C_3\times\cdots C_n$, a product of $n-1$ cyclic groups. Proceeding by induction on the number of terms, we conclude that $G$ is cyclic.

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Luiz Cordeiro
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Induction by n. Suppose the property is true for all $k \lt n$.

We'll prove it for $n$. If $n$ is prime then $G$ is cyclic.

Suppose $n=pq$ where $p,q$ are coprime.

Now let $H_1, H_2$ subgroups of $G$ of orders $p,q$ (we know from Cauchy theorem there are such subgroups). Then, by induction hypothesis, both subgroups are cyclic.

Let $x \in H_1, y \in H_2$ of order $p, q$. Then z=xy will be of order $pq=n$ so $G$ is cyclic.

Now suppose $n=p^k, k \ge 2$ where $p$ is prime.

Then $G$ is a p-group ( group in which every element has order $p^k$ for some $k$). The prove for $G$ cyclic in this case can be found here. Notice that the property "G has at most d elements of order d" implies $G$ has at most $p$ elements of order $p$ hence there is a unique subgroup of order $p$.

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