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Exercise
Let $G$ be a finite group and suppose that $mathbfC$ is a family of subsets of a group $G$ which for a partition of $G$. Further suppose that $gC in mathbfC$ for any $g in G$ and $C in mathbfC$

Prove that $mathbfC$ is a set of cosets of some subgroup of $G$

Solution

So basically I need to show that all elements of $mathbfC$ have same cardinality, $1 in C in mathbfC$ is a group and all elements of $mathbfC$ are of the form $gC$ for some $g$.(Correct me if I am wrong)

Cardinality:
Pick arbitraty $C_i,C_j in mathbfC$. Since $mathbfC$ form a partition they both are non empty. Hence $exists a,b in G$ such that $a in C_i, b in C_j$. By "Further suppose that $gC in mathbfC$ for any $g in G$ and $C in mathbfC$
" we know that $ab^-1 C_r in mathbfC$ and $(ab^-1)b=a in C_r$. Hence it follows that $|C_r|=|C_j|$ because each element of G is contained in exactly one element of $mathbfC$.

Please correct me if I am wrong. Proving that $1 in C in mathbfC$
I am quite sure, however not sure how to prove last part, that is "all elements of $C$ are of the form $gC$ for some $g$."

You need to find or construct the subgroup that produces these alleged cosets. If they were indeed cosets, the part of the partition that contains $1$ must be that subgroup. Then, by the hypotheses of the problem, every left coset of this subgroup must be a part of the partition, which forces the partition to consist only of left cosets. So, all you really need to do is prove that, if $1 in C in mathbfC$, then $C$ is a subgroup.

Suppose $x, y in C$. Note that $1 = x^-1 x in x^-1C in mathbfC$, and given $mathbfC$ is a partition, $x^-1 C = C$. Thus, $x^-1 y in x^-1 C = C$, proving $C$ is a subgroup, as required.

EDIT: By request, here is the full proof:

Suppose $G$ is a group, and $mathbfC$ is a partition of $G$ that is closed under left multiplication by elements of $G$. That is, if $C in mathbfC$ and $g in G$, then $gC in mathbfC$. (Note that the definition of partition requires that $emptyset notin mathbfC$.)

We wish to show that there exists a subgroup $H$ of $G$ such that $mathbfC$ is the set of left cosets of $H$. That is, we must find a subgroup $H$ such that, $gH in mathbfC$ for all $g in G$, and for any $C in mathbfC$, there exists a $g in G$ such that $C = gH$.

Since $mathbfC$ partitions $G$, we have $1 in G = bigcup mathbfC$, hence there must be some $C in mathbfC$ such that $1 in C$. As you and I have both shown, this is a subgroup of $G$. Let $H = C$.

So, since $H in mathbfC$ and $mathbfC$ is closed under left multiplication, we know that $gH in mathbfC$ for all $g in G$. That is, $mathbfC$ contains all left cosets of $gH$.

On the other hand, suppose $C in mathbfC$. We wish to show it is a left coset. Since $C neq emptyset$, choose $g in C$. Then, $g^-1C in mathbfC$, and
$$1 = g^-1 g in g^-1C cap H,$$
so $g^-1 C cap H neq emptyset$. Since these sets are parts in a partition, we must have
$$g^-1 C = H implies gg^-1 C = gH implies C = gH.$$
So, every $C in mathbfC$ is a left coset of $H$, completing the proof.

(We did not need to show that the cardinalities of the parts are the same. This can be deduced from the proven fact that the parts are all left cosets of the same subgroup.)

Let H be the set in C containing e.
Then for any x in H, xH=H( since hH intersect H), which shows that H is a subgroup.
Now all the left cosets of H must be in C, by assumption.
But the all left cosets of H forms a partition already. Thus the set C is the set of all the left coset of H.