Why does this cyclic subgroup have only 4 subgroups? Let the cyclic group have 6 elements and be denoted as $G = \{1, a, a^2, a^3, a^4, a^5\}$ where $a^6 = 1$.
Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $\{1, a^2, a^4\}$ and $\{1, a^3\}$.
Why isnt $\{1, a^5\}$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$? 

There should be an element, $b$ such that $a^3 \cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 \cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated. 
If $a^5 \cdot b = 1$ is true, then $b$ would have to be $a^{-5}$ or $a^{10}$, where it is explicitly stated that $a^{10} = 1$ as well.
Is my thought process correct?
 A: $[1,a^5] $ is not a subgroup because $a^5\cdot a^5=a^4$ which is not in the set $[1,a^5]$
But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.
A: $\lbrace 1, a^5 \rbrace$ is not a subgroup because 
$$a^5 . a^5 = a^4$$
is not an element of $\lbrace 1, a^5 \rbrace$. So $\lbrace 1, a^5 \rbrace$ is not stable for the intern law of the group.
A: Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.
It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $\langle a^k \rangle = \langle a^{gcd(n,k)} \rangle$. Where $\langle a \rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$. 
A: Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

To address your misunderstanding: if $g\in H$ for some $H\le G$, then all powers of $g$ are in $H$.
