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I can see how it is indeed possible that $G$ has a proper subgroup i.e. possibly $n = km$ and there is an element of order $k$ that generates a cyclic proper subgroup, but I am trying to see why this must necessarily be the case. My argument goes as follows:

Let $G$ be a group of order $n$. Since $n$ is composite, we may write $n = km$ for $k,m \in\Bbb Z$. If $G$ is cyclic, let the element that generates $G$ be $a$. Now consider $\langle a^k \rangle$ for $k$ the above divisor of $n$. This subgroup is nontrivial, and contains $m < n$ elements by Lagrange's Theorem, so we are done.

Else, if $G$ is not cyclic, then $n$ composite implies $n \ge 4$, thus there are at least 4 elements in $G$, none of which has order $n$. Consider an element $b \in G$, $b \ne e$. Let the order of $b$ be $\ell < n$. Then $b$ generates a cyclic subgroup of order $\ell$ which must divide $n$, and since $\ell < n$, we know that $\langle b \rangle \subsetneq G$. Thus $G$ has a proper subgroup. $\Box$


Any feedback about the above proof is appreciated!

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It's ok but too complicated. If $G$ has no proper subgroups take $x\in G$, $x\neq e$. Then $<x>=G$, so $G$ is cyclic, and finite cyclic groups, as you observed, have subgroups for any divisor of their order. –  user26857 Dec 4 '12 at 0:51
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up vote 2 down vote accepted

If $G$ has order $n$ which is composite, then $n$ is divisible by some prime $p < n$. By Cauchy's theorem, $G$ has an element of order $p$. This element generates a proper subgroup of order $p$.

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How can I see this is true without Cauchy's Theorem. For example, how would I know that a group of even order must have an element of order 2? –  Zvpunry Dec 4 '12 at 1:01
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In a group of even order, pair off every element with its inverse. That is an even number of elements. Plus the identity which brings the total to an odd number. Therefore, there must be at least one element left whose inverse is itself, i.e., an element of order 2. –  user44441 Dec 4 '12 at 2:31
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