Without using Cauchy's theorem: If $G$ an abelian group of order $10$ contains an element of order $5$, show that $G$ must be a cyclic group.

Question

If $G$ an abelian group of order $10$ contains an element of order $5$, without using Cauchy's theorem, show that $G$ must be a cyclic group.

In my book, there is a hints that $G$ has an element of order $2$. But without using Cauchy's theorem, here how to show this?

Please help me to solve the remaining part of the problem.

Answer 1

The possible orders of non-trivial elements aer $2,5,10$. If we have $10$ we are done. Now suppose there are no elements of order $2$, then every element has order $5$. Take $x$ one such element and consider $G/<x>$ ($G$ is abelian!). Now this group has order $2$, so it must be isomorphic to $\mathbb{Z}/2\mathbb{Z}$. This means that for all $y\in G$, $y^2\in <x>$. Take $z$ such that $z\not\in <x>$. Then $z^2\in <x>$, so $z^2=x^a$ for some $0\leq a\leq 4$. If $a=0$ we found our element of order $2$, whereas if $a\neq 0$ we have an element of order $10$.

Answer 2

You can do this by a counting argument. Let $ x$ be the element of order $5$, and let $y\in G\setminus \langle x\rangle$. If $ y$ has order $ 5$, then $\langle x\rangle \cap\langle y \rangle= \{e\}$, since if $y^k=x^j$ for some $k$, then by raising to the power of $k^{-1}\pmod 5$, we get $y\in \langle x \rangle$.

But there are only $9$ elements in $\langle x\rangle \cup\langle y \rangle$ and $G$ has $10$ elements so there is one element $z$ left. By the same argument as above $z$ cannot have order $5$.

So there exists an element $z$ of order $m\ne 5$.

Answer 3

Suppose we have an element $x$ with order 5. Then, we have $<x> = \{e,x,x^2,x^3,x^4\} \subset G$, and we see that the inverse of each element of this set is contained in the set, since $e^{-1} = e$, $x^{-1} = x^4$, and $(x^2)^{-1} = x^3$. This means that in the remaining 5 elements of $G$, say, $G \setminus <x> = \{y_1,y_2,y_3,y_4,y_5\}$, the inverses of these elements must also be in $G \setminus <x>$. This comes directly from the fact that elements of a group have unique inverses (why?), so the inverses of any of $y_1,\ldots,y_5$ cannot be a member of $<x>$. Then, we see that since we have an odd number of elements in $G \setminus <x>$, one of its elements must be its own inverse. Thus, we have proven there must exist an element of $G$ with order 2.

Let us call this element of order 2 $k$ (which is not a power of $x$). Clearly, since any product of the form $x^ik \notin <x>$ but $x^ik \in G$ due to closure of a group, we must have $G \setminus <x> = \{k,xk,x^2k,x^3k,x^4k\}$. Now, $G$ is starting to look just like $D_{5}$, except for the fact that $G$ is abelian!

Using this fact, we have $(x^2k)^2 = x^2kx^2k = x^2x^2kk = x^4$, $(x^2k)^3 = x^4x^2k = xk$, and so on. What you will find is that $<x^2k> = G$, and therefore $G$ is cyclic.