Suppose $G$ is a group with exactly $8$ elements of order $3$. How many subgroups of order $3$ does $G$ have? Suppose $G$ is a group with exactly $8$ elements of order $3$. How many subgroups of order $3$ does $G$ have?
 A: Note that, if $a \in G$ is an element of order 3, then so is $a^2 = a^{-1}$.
Besides every subgroup $H = \{ a,a^{-1}, 1\} \leq G$ of order 3 is cyclic and is generated both by $a$ and $a^{-1}$.
This means that we can enumerate the eight elements of order 3 in this way: $a, a^{-1}$,$ b, b^{-1}$,$ c, c^{-1}$, $d , d^{-1}$ and the four subgroups of order 3 are $A= \{ a,a^{-1}, 1\}$, $B= \{ b,b^{-1}, 1\}$, $C= \{ c,c^{-1}, 1\}$ and $D= \{ d,d^{-1}, 1\}$.
A: If $a\in G$ has order 3, $a^2$ also has order 3. So, if $a,b,c,d$ and their squares have order 3, all subgroups containing identity,element and square of element is a subgroup of order 3.
A: If $a\in G$ has order $3$ then the subgroup generated by $a$ is given by $\langle a \rangle = \{1, a, a^2\}$.
Note that $a^2$ also generates $\langle a \rangle$ and is not equal to $a$ and hence each subgroup $\langle a \rangle$ contains two elements of order $3$.
Next note that if $a \in \langle b \rangle$ where $a,b$ have order three then $a = b$ or $a=b^2$, that is, any two such subgroups are mutually disjoint.
Since in each group there are exactly two elements of order $3$ we have hereby found four subgroups of order three.
Since a subgroup of order $3$ must be cyclic (because $3$ is prime) these are the only subgroups of order $3$ hence there are exactly four subgroups of order $3$.
