How many elements of order $7$ are there in a group of order $28$ without Sylow's theorem?

Question

How many elements of order $7$ are there in a group of order $28~?$

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I need to prove this result without using the Sylow's Theorem.By Sylow's Theorem it has only one subgroup and the answer becomes $~6$. But in the book the problem is in the Cauchy's Theorem chapter and I want to prove it by this only.Can someone help me please.

Answer 1

By Cauchy's theorem the group, say $G$, has an element $g$ of order $7$. To the contrary, assume $G$ has an element $h$ of order $7$ not contained in $\langle g \rangle$. Then, $\langle g \rangle \langle h\rangle$ has size $49 > 28$ (why?), a contradiction.