Ttykuu

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?

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ 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.

$lbrace 1, a^5 rbrace$ 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.

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 $a$.

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

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