I have been asked a question form my lecturer and he said I am wrong.
Assume you have a group $G$, known that there is exactly 28 elements in order 5, how many sub-groups $G$ has in order 5?
My thoughts and logic is that is known that an element in a group such that its order is $k$ therefore it generates sub-group in order $k$ aswell.
Also, it is known that in every group (and particularly sub-group) every element has invert element such that they have the same order,
therefore I can conclude that if there is 28 elements order 5, divide by $2 \rightarrow 14$ sub-groups.
Where am I wrong?
Since $5$ is prime number, all subgroups with order $5$ are cyclic groups. Let
$$H_1=\langle g_1\rangle=\{e,g_1,g_1^2,g_1^3,g_1^4 \}=\langle g_1^2\rangle=\langle g_1^3\rangle=\langle g_1^4\rangle,$$
hence there are four generators for $H_1$. Suppose there is another subgroup $H_2$ with order $5$, then
$$H_2=\langle g_2\rangle=\langle g_2^2\rangle=\langle g_2^3\rangle=\langle g_2^4\rangle.$$
Because $H_1\neq H_2$, it means none of their generators is contained in another group. And for a subgroup with order $5$, all of its elements have order $5$, except for the unit element $e$.
Therefore, there is a total of $28/4=7$ subgroups of order $5$.
Another approach,
Suppose there is a total of $k$ subgroups of order $5$. Assume $k>7$, then there are at least $4\cdot k\ge32$ elements with order $5$, which contradicts with the fact there is a totall of $28$ elements of order $5$.
Assume $k<7$, then these $k$ subgroups $H_1,\dots, H_k$ supply at most $4\cdot 6=24$ elements of order $5$, but we are told there are totally $28$ elements of order $5$, which means exist some elements $g\notin \cup_{i=1}^k H_i$ and $|g|=5$, so we get one more subgroup $\langle g\rangle$ of order $5$, which contradicts with the assumption there is a total of $k$ subgroups of order $5$ (because now we get $k+1$ subgroups).