A cyclic group is defined as a group which can be generated by a single element.
For example, let $G = \Bbb Z/5\Bbb Z = \lbrace 0, 1, 2, 3, 4\rbrace$ with addition. Then
$$\begin{align}
1 &\equiv 1 \bmod 5\\
1 + 1 &\equiv 2 \bmod 5\\
1+1+1 &\equiv 3 \bmod 5\\
1+1+1+1 &\equiv 4 \bmod 5\\
1+1+1+1+1 &\equiv 0 \bmod 5
\end{align}$$
and since these are all of the elements of $G$, we conclude that this group is cyclic, and is generated by $1$, which can be written $G = \langle 1\rangle$ (although this requires some context, simply writing $\langle 1\rangle$ doesn't give enough information).
The way the elements cycle comes from the way the group $\Bbb Z/n\Bbb Z$ is defined. The "number" in the set $\lbrace 0, 1, 2, \dots, n-1\rbrace$ actually refers to the remainder when an integer is divided by $n$. For example, in the case $n = 5$, we see that $23 = 4 \times 5 + 3$, so $23 \equiv 3 \bmod 5$. Since the only possible candidates for remainders after dividing by $5$ are the numbers $0, 1, 2, 3, 4$ (convince yourself of this; for example, if I wrote $23 = 3 \times 5 + 8$, I could take another $5$ from the $8$ and have $23 = 3 \times 5 + 5+ 3 = 4\times 5 + 3$), these are the elements of the group.
Cyclic groups are abelian. Let $h\in H$, $g\in G$ for $H\le G$, abelian $G$. Then $$ghg^{-1}=gg^{-1}h=h\in H.$$ Thus $H\unlhd G$.
Best Answer
Let $\psi(d)$ denote the number of elements of order $d$ in $\mathbb Z_p^*$.
Note that if $a\in\mathbb Z_p^*$ has order $d$, then $a^r$ has order $d$ if and only if $\gcd(r,d)=1$.
Thus are precisely $\phi(d)$ elements of the form of $a^r$ which have order $d$.
It follows that $\phi(d)$ divides $\psi(d)$ for all $d$.
Using (1) we want to show that $\psi(d)$ is either $0$ or equal to $\phi(d)$ for all $d|(p-1)$.
Assume on the contrary that there is some $d$ dividing $p-1$ such that $\psi(d)>\phi(d)$.
Let $a$ be an element of order $d$.
Out of the $d$ elements $a,a^2,\ldots,a^d$, precisely $\phi(d)$ have order $d$.
By our assumption, there is an element $b$ of order $d$ which is not equal to any of the $a^i$'s.
But note that $b$ and each $a^i$ satisfies $x^d\equiv 1\pmod{p}$.
This means that the congruence $x^d\equiv 1\pmod{p}$ has more than $d$ solutions, contrary to (1).
So we have $\psi(d)=0$ or $\phi(d)$.
Now note that $\sum_{d|(p-1)}\psi(d)=p-1$.
This is because each element of $\mathbb Z_p^*$ has order $d$ for some $d|p-1$.
From our earlier inference we also have $\sum_{d|p-1}\psi(d)\leq \sum_{d|p-1}\phi(d)$.
The RHS is equal $p-1$ by (2).
Therefore we cannot have $\psi(d)=0$ for any $d|p-1$ and hence $\psi(p-1)>0$, proving there is an element of order $p-1$ (and hence precisely $\phi(p-1)$ such elements).