a) Let $G_1$ and $G_2$ be groups, each with exactly one element. Show $G_1\cong G_2$.
b) Let $G_1$ and $G_2$ be groups, each with two elements. Show $G_1\cong G_2$. Does this statement also hold for three elements? And for four?
a) Let $g_1$ be the element in $G_1$, and $g_2$ the element in $G_2$. Let $f\colon G_1\to G_2$ be defined as f(g_1)=g_2. We know that $g_1$ and $g_2$ are the neutral elements in $G_1$ and $G_2$ respectively. So $f(g_1g_1)=f(g_1)=g_2=g_2g_2=f(g_1)f(g_1)$.
b) The same kind of reasoning can be applied for the case of 2 elements (where we have a neutral element, and an element $a\in g_1$ which is its own inverse). For three elements, we could consider $e,a,b\in g_1$, with either $a^{-1}=b$, or $a^2=b^2=e$. This still seems doable for me to check. However, how am I supposed to consider the case of 4 elements? Should I still list all possibilities and check if there exists an isomorphism? Or can this be done more quickly?
EDIT
Based on the comments below and some help from the chat:
There is only one group with three elements, because we can't have that $aa=e$, otherwise we would have that $aa=ae=e$, which would mean that $a=b$.
I've considered the groups $\mathbb Z/4\mathbb Z$ and $(\mathbb Z/2\mathbb Z)\times(\mathbb Z/2\mathbb Z)$. I won't write out their Cayley tables, but it is clear that these are not isomorphic groups. I think the most telling property is that for each $\overline x\in(\mathbb Z/2\mathbb Z)\times(\mathbb Z/2\mathbb Z)$, we have that $\overline x+\overline x=e$, however, for $\mathbb Z/4\mathbb Z$ this is not the case (consider: $\overline 1+\overline 1=\overline 2\neq\overline 0$). Back to the Cayley table: these two groups are never the same on the diagonal, no matter how you order their elements.
Or you can use more general results. Let $G$ be a group of prime order $p$. Then $G\simeq\mathbb{Z}_p$. Indeed, let $g\in G$ be a nontrivial element. Define a homomorphism
$$f:\mathbb{Z}\to G$$ $$f(n)=g^n$$
By the Lagrange's theorem the order of the image $|f(\mathbb{Z})|$ divides $|G|$. Since $|G|$ is prime and $|f(\mathbb{Z})|>1$ because $g$ is nontrivial, then $f(\mathbb{Z})=G$. Obviously $\ker(f)=p\mathbb{Z}$ (because $g$ is of order $p$) and by the first isomorphism theorem
$$G\simeq \mathbb{Z}/p\mathbb{Z}\simeq\mathbb{Z}_p$$
It applies to your case since $2$ and $3$ are primes.
For the case $4$ take two groups $G=\mathbb{Z}_4$ and $H=\mathbb{Z}_2\oplus \mathbb{Z}_2$. The both have $4$ elements but $G$ has an element of order $4$ while all (nontrivial) elements in $H$ are of order $2$. Thus they cannot be isomorphic.