Show group can't have 12 elements of order 7 Let $G$ be a group show that $G$ can't have 12 elements of order 7. 
my try: 
If there is $a \in G$ such that $o(a)=7$ then $<a>$ has 6 distinct elements of order 7 then suppose $<a>$ is not normal in $G$ then there is element $g \in G$ such that $gag^{-1} \not\in <a>$ so $o(gag^{-1})=7$ and it posses $6$ elements of order 7 then $<a> \cap <gag^{-1}> = \{e \}$ so it gives us $12$ elements so far and I don't know how to contradict this 
 A: You've correctly seen that you can have two distinct subgroups of order $7$, and these would contain $12$ elements of order $7$ between them.  So to continue, you must now show that in this situation, there are always more elements of order $7$.
Here's a hint for how to continue:
Call the generators of the two subgroups $a$ and $b$, and consider $bab^{-1}$.  Now split the argument into three cases:
1) $bab^{-1} \in \langle b\rangle$.  Show this contradicts the assumption $\langle a\rangle \neq \langle b\rangle$.
2) $bab^{-1} \in \langle a\rangle$.  In this case, you know that $bab^{-1} = a^n$.  Now carefully consider $ab$.
3) $bab^{-1}$ is in neither $\langle a\rangle$ nor $\langle b\rangle$.  Now what can you say about $bab^{-1}$?
A: I think the answer by MartianInvader will work, but let's see if we can do it another way that doesn't require examining multiple cases.
As you showed already, if $G$ has exactly $12$ elements of order $7$, then it has exactly two subgroups of order $7$, call them $A$ and $B$.
Conjugation by any element $g$ is an automorphism on $G$, so it must either fix (normalize) both $A$ and $B$ or it must swap them. Consider conjugation by an element $g \in A$: this must normalize $A$, and therefore it must also normalize $B$. In other words, every element in $A$ normalizes $B$, so $A \leq N_G(B)$. Since of course we also have $B \lhd N_G(B)$, this means that $AB$ is a subgroup. The order of $AB$ is
$$|AB| = \frac{|A||B|}{|A \cap B|} = |A||B| = 7^2$$
By Lagrange's theorem, every element of $AB$ must have order $1$, $7$, or $7^2$. Only the identity has order $1$, and since there are only $12$ elements with order $7$, then the rest must have order $7^2$, which means that $AB$ is cyclic. But this is impossible, because a cyclic group has exactly one subgroup of each possible order, whereas $AB$ has two subgroups of order $7$, namely $A$ and $B$.
