$G$ a group of order 210.
Part 1 If $G$ is commutative show it is cyclic.
Consider the prime factorization $210=(2)(3)(5)(7)$. By Sylow's theory we know that Sylow subgroups exist of order 2, 3, 5, and 7 in $G$. Let $P_i$ be a Sylow subgroup of order $i$, for $i \in \{2,3,5,7\}$. Each one of these subgroups is a cyclic group because the order is prime.
Consider the the set $P_i P_j :=\{qh|q \in P_i, h \in P_j \}$. $P_iP_j$ is a subgroup since it is closed under products and inverses (proof omitted). $|P_iP_j|=ij$ by the simple combinatorics of the number of unique ways of choosing an element from a set $i$ elements and another from a set of $j$ elements, regardless of order (commutative property).
Finally I want to show that $P_i P_j$ is cyclic because it has an element of order $ij$, and so this element generates the whole group. For $qh \in P_iP_j$ such that $q,h \ne 1$, and assume that $(qh)^n=1$. Then $q^nh^n=1$, by the commutative property. This happens if and only if $q^n=1$ and $h^n=1$, so $n$ must be a multiple of $i$ and also a multiple of $j$. Since, $i$ and $j$ are coprime, $ij$ is the smallest number that is a multiple of both $i$ and $j$, and as such is the order of $qh$.
This shows that $P_iP_j$ is cyclic. A very similar argument can be used to show that $P_2 P_3 P_5 P_7$ defined in the same way is cyclic, because 2,3,5, and 7 are coprime.
Part 2 If $G$ is any group show it contains a subgroup of index 2.
For this part, I use the lemma that if $A$ and $B$ are finite subgroups of a group, then $$|AB|=\frac{|A||B|}{|A \cap B|}$$ with emphasis that for any $P_i, P_j$, where $i,j \in \{2,3,5,7\}$, are the Sylow subgroups referenced above, then $|P_i \cap P_j| = \{1\}$. If not, then $P_j$ would contain an element of order $i$, and thats a contradiction since each $P_j$ is a cyclic group. From here, it is straight-forward. $$|P_5 P_7|=\frac{(5)(7)}{1}=35$$ Now, I claim also that $|P_3 \cap P_5P_7|=\{1\}$. If not then $P_5P_7$ would contain an element of order 3, which it cannot, because by Lagrange, 3 does not divide 35. Et voila, $$|P_3 P_5 P_7|=\frac{|P_3||P_5 P_7|}{|P_3 \cap P_5 P_7|}=\frac{(3)(35)}{1}=105$$ The index of a subgroup of order 105 is 2.
Here is a simpler proof of part 1: By Cauchy's theorem, there are elements of order $2,3,5,7$. Then the product of these elements has order $210$.