Number of elements of order $7$ in a group 
If $G$ has at least $7$ elements of order $7$, then prove that $G$ has at least $12$ elements of order $7$.

I know that an element having order $7$ means that for any $a$ in $G$, $a^7=e$ where $e$ is the identity element. The inverses also have same order, so I thought maybe of the $7$ elements, $6$ had their inverse as well. However, then this would just mean there had to be $8$ elements, not $12$ for the $7$th element to have it's inverse. 
 A: If there is an element with order 7, we have a subgroup of the group which is isomorphic to the cyclic group with 7 elements, namely the subgroup generated by the element of order 7.
All elements in this subgroup also have order seven (by Lagrange's theorem) excepting the identity element. So there exists another element with order seven out of the subgroup. This element also generates a subgroup of order 7 which meets the other subgroup only in the identity element (because intersections of subgroups are subgroups and by Lagrange a group of order seven has only the trivial subgroups), so we get 5 more elements of order seven, namely the other elements in the second subgroup.
So all in all we have at least 12 elements of order 7.
A: Let $a$ be an element of order 7 in $G$. Then $\{a,a^2,\dots a^6\}$ are all elements of order 7. Indeed this set (along with the identity $e$) comprises a subgroup of $G$ so all $a^k$ ($k=1,\dots,6$) are order $7$ (by primality of 7).
By assumption there exists $b\neq a^k$. Again, by considering the subgroup generated by $b$: $\{b,\dots,b^6,e\}$, we observe that $b^j$ ($j=1,\dots,6$) are all elements of order 7, each of which is distinct from each $a^k$.
EDIT: 
To understand why each $b^j$ is distinct from each $a^k$, suppose to the contrary that $b^j=a^k$ for some $j,k\in \{1,\dots,6\}$. Since each $b^j$ is a generator of the subgroup $\langle b\rangle$ we know that $(b^j)^h=b$ for some $1\leq h \leq 6$. However, this implies that $b=(b^j)^h=(a^k)^h$ contradicting the choice of $b\neq a^k$ for all $k$.
A: Let $a$ be an element of order $7$ in $G$, i.e., $a^7=e$, where $e$ is the identity of $G$. Now if we take the powers of $a$ we form the cyclic group of order $7$:
$$\{a,\, a^2,\,  a^3,\,  a^4,\,  a^5,\,  a^6,\,  a^7=e \}=\langle a\rangle\cong C_7$$
Now, since $C_7$ has an order that is a prime number, namely $7$, there can be no proper subgroups of $C_7$, i.e., no subgroup between the trivial group and $C_7$, since by Lagrange's Theorem the order of any subgroup of $C_7$ has to divide $|C_7|=7$. Hence each of $a^k$ for $1\le k\le 6$ are distinct elements of $G$, which implies each has order  $7$ (were this not the case and $a^i=a^j$ for some $1\le i<j\le 6$ then this $a^j$ would be the generator of a subgroup $\langle a^j\rangle$ whose order divided $7$, which is impossible by Lagrange's Theorem).
Since $G$ has at least $7$ elements of order $7$, and we have accounted for $6$ of them with our construction of $C_7\cong\langle a\rangle$, there must be another copy of $C_7$ in $G$ with generator $b$, where any power of $b$ excepting the identity, is not equal to any power of $a$, and with each power of $b$ distinct as in the case for $C_7$ generated by $a$: 
$$\{b,\, b^2,\,  b^3,\,  b^4,\,  b^5,\,  b^6,\,  b^7=e \}=\langle b\rangle\cong C_7$$
This gives at least $6+6=12$ elements of order $7$ in $G$.
