Which of the following conditions imply that $G$ is abelian? Let $G$ be a group of order $n.$ Which of the following conditions imply
that $G$ is abelian?
a. $n = 15.$
b. $n = 21.$
c. $n = 36.$
I can see $a$ is true (since a cyclic group of order $3\times 5$) and $b$ is false (by considering $D_{36}$). But I'm skeptical about my solution for $b$ since the answer says group of order $21$ is not necessarily abelian even though 

if I consider $H,K$ to be the sylow 3 and sylow 7 subgroup then $G\simeq H\times K.$ Since $H,K$ are abelian so is $G.$ 

I don't know where did I go wrong?
 A: For a concrete example of a non-abelian group of order $21$, let us first consider a broader class of groups, namely those of affine transformations of a finite field.
So let $F$ be a finite field of order $q$ and consider the set of maps $F\to F$ of the form $x\mapsto ax + b$ with $a\neq 0$. It is easy to check that this set forms a group under composition (since these maps are bijective), and this group has order $q(q-1)$.
If $G$ is a subgroup of the multiplicative group of $F$, then restricting to those maps with $a\in G$ gives a new group, with order $q|G|$.
For the specific example of order $21$, we can take the finite field of order $7$ (so $\mathbb{Z}/7\mathbb{Z}$) and the subgroup to be $\{1,2,4\}$. You should then check that the resulting group is not abelian.
The above class of groups is a great way to construct examples of non-abelian groups of various orders.
A: In a group of order $\;15=3\cdot 5\;$ , both Sylow subgroups must be normal (and thus unique) by direct application of Sylow theorems, and this means the group is a direct product of these two subgroups, each of which is abelian (even cyclic) and thus the whole group also is abelian (and cyclic, in fact) .
In the other two cases we can construct a non-abelian group of that order using semidirect products. but since you haven't yet covered this I won't go further.
A: You can construct the (non-abelian) semidirect product of $C_7$ and $C_3$ as follows: We view $C_7$ as the additive integers $\text{mod}7$ and $C_3$ as the subgroup of $\Bbb Z_7^*$ of order $3$, this is the multiplicative group $\{1, 2, 4\}$. Then $G=\Bbb Z_7\times \{1, 2, 4\}$ and multiplication is defined by $$(a,b)*(c,d)=(ad+c,bd).$$ Check that the multiplication is associative, then the neutral element is $(0,1)$ and inverses are given by the rule $(a,b)^{-1}=(-ab^{-1},b^{-1})$.
A: I wonder you already have an answer for some part of your Question :

If I consider $H,K$ to be the sylow 3 and sylow 7 subgroup then $G\cong H\times K$. Since $H,K$ are abelian so is $G$.

i remember you posed similar question asking to verify a proof that "any group of order $85$ is cyclic".
to show $G\cong H\times K$ it is not enough to show $H\leq G$, $K\leq G$ and $|H\times K|=|G|$
you need to prove that $H\unlhd G$ and $K\unlhd G$
In general it is not true that any sylow-$3$-subgroup in a group of order $21$ is normal (you please check it).
So, what you say i.e., $G\cong H\times K$ is not true in general.
So, there may possibly exist a nonabelian group of order $21$.
I hope you are happy now at least that you came out from a wrong notion.
Now expecting the possibility of existence of non abelian group : 

I would suggest you to make a copy of (more generalized) group of order $21$ in a way similar to that you construct a dihedral group. (this is what comes to my mind if i want to construct non abelian group if i am sure there does exist a possible non abelian group. though i fail most of the times (:P) i feel one should not fear of getting defeated)

So, what we do in dihedral group $D_{2n}$ is take an element $r$ of order $n$ and an element $s$ of order $2$ and generate the group with an extra condition that $srs^{-1}=r^{-1}$.
With same motivation, as $21=3.7$ i take an element of order $7$ and an element of order $3$ and write very similar relation as that of  $srs^{-1}=r^{-1}$.
I know there exists an element of order $7$ (thanks to cauchy theorem and sylow(??)) say $a$ and an element of order $3$ (thanks to cauchy theorem and sylow(??)) say $b$.
we have $a,b\in G$ with $'a'$ generator of sylow $7$ subgroup and $'b'$ a generator of sylow $3$ subgroup. 
As we know that any sylow $7$ subgroup is normal, $bab^{-1}\in \langle a\rangle$ (??)
So, $bab^{-1}$ has to be one of $1,a,a^2,a^3,a^4,a^5,a^6$
one see that $bab^{-1}\neq 1$ in which case $a=1$ and $bab^{-1}\neq a$ (in which case $ab=ba$ and so $G$ is abelian and we do not want that).
So the next option is $bab^{-1}=a^2$, I do not see any problem with this and i see that i have exactly $21$ elements (??) in $\{\big<a,b\big> : |a|=7, |b|=3 ,bab^{-1}=a^2\}$.
So, I have got a non abelian group of order $21$ with 

$G=\{\big<a,b\big> : |a|=7, |b|=3 ,bab^{-1}=a^2\}$

please try to check what would happen if $bab^{-1}=a^i$ for $3\leq i\leq 6$ (which i have not checked out)
P.S : I tried my best to use nothing more than cauchy/sylow theorem as you said you are not familiar with semi direct product. please let me know if you see any traces of that as i also want to write it down more elementarily. 
