Proving $G\cong \Bbb{Z}_p$ if $|G|$ is prime. For a group $G$, if $|G|$ is prime, then I have to prove $G\cong \Bbb{Z}_p$.
Take any element $g\in G$. As $G$ cannot have proper subgroups, and as it also has finite order, $|g|=p$. Hence, $G=\{1,g,g^2,\dots,g^{p-1}\}$. We know $\Bbb{Z}_p=\{0,1,2,\dots,p-1\}$. Hence, $f:G\to\Bbb{Z}_p$ defined by $f(g^r)=r$ for any $r\in\Bbb{Z}_p$ is a homomorphism. 
If we can now prove that the kernel of this mapping is $\{1\}$, then we're done. Clearly, $1$ is the only element that maps to $0$ in the above homomorphism. 


*

*Is the above reasoning correct?

*What are some other ways of proving isomorphism between the two groups? Is proving that the kernel is composed of only one element the only way? I read somewhere that if provided you have proven $f$ is a homomorphism, and that $h$ is another mapping from $\Bbb{Z}_p$ to $G$, then proving $fg$ is injective and $gf$ is surjective is sufficient to prove isomorphism. But what are the properties of $g$? 

 A: As to your point 2, your question screams for an application of the first isomorphism theorem.
Define 
$$
f : \mathbb{Z} \to G, \qquad n \mapsto g^n,
$$
where $1 \ne g \in G$. Because of the rule $g^{n+m} = g^{n} g^{m}$, this is a group homomorphism. 
Since $G$ has no proper, nontrivial subgroups, and the image of $f$ contains $g \ne 1$, it follows that $f$ is surjective, and thus
$$
\mathbb{Z} / \ker(f) \cong G,
$$
so that $p = \lvert G \rvert = \lvert \mathbb{Z} : \ker(f) \rvert$.
Thus $\ker(f)$ is the unique subgroup of $\mathbb{Z}$ of index $p$, that is, $\ker(f) = p \mathbb{Z}$.
A: Suppose $|G|=p$, a prime.
Then $|G| > 1$, so there is a non-identity element.
Let $a$ be a non identity element of $G$.
Consider the subgroup generated by $a$:  $\langle a\rangle$
Since $a$ is not $1$,  $a^0 = 1$ and $a^1$ are two distinct elements of $\langle a\rangle$
Thus $|\langle a\rangle |> 1$ ... call this fact (*).
Now by Lagrange's theorem, $|\langle a\rangle|$ must divide $|G|=p$, a prime
The only positive integers that divide prime $p$, are $1$ and $p$
By fact (*), $|\langle a\rangle|$ cannot be $1$, so we must have $|\langle a\rangle| = p$.
Now all cyclic groups of order $p$ are isomorphic to $\mathbb{Z}_p$, so you are done.
If you cannot use this fact, then exhibit an isomorphism $f: \langle a\rangle \rightarrow \mathbb{Z}_p$ by
$f(a^n) = n$
Proof that it is a homomorphism:
$f(a^m \cdot a^n) = f(a^{(m+n)} ) = m + n = f(a^m) + f(a^n)$
Proof that it is an isomorphism (that is, prove $f$ is bijective):
Let's exhibit an inverse of $f$, say $g$:
$g(n) = a^n$
proof it is an inverse:
$f(g(n)) = f(a^n) = n$ , and this is true for all $n$ in $\mathbb{Z}_p$
$g(f(a^n)) = g(n) = a^n$, and this is true for all $a^n$ in $\langle a\rangle$
Thus $f$ is an isomorphism from $\langle a\rangle = |G|$ onto $\mathbb{Z}_p$.
