Group of order $7^2 \cdot11^2\cdot19$ is abelian. I want to show any group of order $7^2 \cdot11^2\cdot19$ is Abelian, I know that $n_7=1$ and $n_{11}=1$ or $7\cdot19$ and $n_{19}=1,7\cdot11$ or $7^2\cdot11^2$. But I don't know where I should go from here.
 A: Let $G$ be a group of order $7^2 \cdot 11^2 \cdot 19$. Let $H, K, L$ be a $7$-sylow, $11$-sylow, $19$-sylow of $G$ respectively, then $H \lhd G$. 


$G$ have a normal subgroup of order $7^2\cdot 11^2$.

Note that $H K$ is a subgroup of index $19$, so we a permutation representation on these $19$ cosets (via left multiplication): 
$$\varphi: G\to S_{19}$$ 
this cannot be an embedding because $11^2 \nmid 19!$. $\varphi$ is neither trivial because the action is transitive. Also note that $H \subset \ker \varphi$ because of normality. Hence
$$|\ker \varphi| = 7^2 \quad \text{ or } \quad 7^2\cdot 11 \quad  \text{ or } \quad 7^2\cdot 11^2 \quad  \text{ or } \quad 7^2\cdot 19 \quad  \text{ or } \quad 7^2\cdot 11 \cdot 19$$
with corresponding size of image:
$$|\text{im } \varphi| = 11^2\cdot 19 \quad \text{ or } \quad 11\cdot 19 \quad  \text{ or } \quad 19 \quad  \text{ or } \quad 11^2 \quad  \text{ or } \quad 11$$
since the action is transitive and image is a subgroup of $S_{19}$, we can eliminate all except the case $|\text{im } \varphi| = 19$, with corresponding $|\ker \varphi| = 7^2\cdot 11^2$.


$K \lhd G$.

Let $F\lhd G$ be the normal subgroup of order $7^2 \cdot 11^2 $ obtained above. Then it is easy to verify $F$ is abelian and $H \leq F$. Let $Q$ be the $11$-sylow in $F$, then $F = H Q$. For any $g\in G$,
$$g HQ = HQ g \implies HgQ = HQg \implies gQg^{-1} \subset HQ$$
hence all $11$-sylow of $G$ is contained in $HQ$. However, $HQ$ is abelian, so there is in fact only one $11$-sylow.


$G$ is abelian.

Let a set of geneators of $K$ be $\{k_i\}$, of $H$ be $\{h_i\}$. Let a geneator of $L$ be $l$. If $k\in \{k_i \}, h\in \{h_i \}$, then for some integers $r,j$:
$$lkl^{-1} = k^j \implies j^{19} \equiv 1 \pmod{11}$$
$$lhl^{-1} = h^r \implies r^{19} \equiv 1 \pmod{7}$$
The only possible solution is $r\equiv 1, j\equiv 1$. Hence $G$ is abelian. 
A: A slightly different approach.
1) use Sylow's theorems to show that every group of order $11^2.19$ is abelian.
2) (as already done by the OP) use Sylow's theorem to show that the 7-Sylow $H$ in $G$ is normal. 
3) Since the automorphism group of $H$ has order $6\times 7$ or $6^2\times 7\times 8$, the action of the quotient $G/H$ on $H$ is trivial, so $H$ is central.
4) For $p\in\{11,19\}$, let $M$ be a $p$-Sylow in $G$. Then the image of $M$ in $G/H$ is the unique $p$-Sylow of the abelian group $G/H$. The inverse image in $G$ of the latter is the direct product $H\times M$. So $M$ is uniquely defined. Hence for both $p=11$ and $p=19$, $G$ has a unique $p$-Sylow. So $G$ is direct product of its Sylow subgroups, and since the exponents of primes are $\le 2$ it follows that $G$ is abelian.

Additional remarks. This method works with no change for any group of order $q_1^2q_2^2q_3$ when $q_1,q_2,q_3$ are distinct primes such that 
(a) $q_3$ does not divide $q_2-1$ or $q_2+1$, and $q_2$ does not divide $q_3-1$
(b) none of $q_2,q_3,q_2^2,q_2q_3,q_2^2q_3$ is equal to 1 modulo $q_1$ (so (2) adapts). 
(c) none of $q_2,q_3$ divides $q_1-1$ or $q_1+1$ (so that (3) adapts)
[Note that (c) is a necessary condition, since otherwise there exists a non-abelian group of order $q_1^2q_i$ for some $i\in\{2,3\}$. Similarly (a) is a necessary condition. In (b) it's also necessary that $q_2,q_3,q_2^2$ are $\neq 1$ modulo $q_1$, but (b) is not necessary, see below.]
Using that groups of odd order are solvable, optimal conditions are as follows:
Fact: let $q_1,q_2,q_3$ be distinct primes. Then every group of order $q_1^2q_2^2q_3$ is abelian if and only the following hold
(A) $q_1,q_2$ do not divide $q_3-1$
(B) $q_1,q_3$ do not divide $q_2\pm 1$
(C) $q_2,q_3$ do not divide $q_1\pm 1$.
Note that these conditions imply that all $q_i$ are $\ge 5$. An example not covered by the previous case is $(5,7,13)$ because $q_2q_3$ equals 1 modulo $q_1$. Still since $q_1,q_2$ play the same role, and $(7,5,13)$ is covered. Anyway, $(5,7,73)$ satisfies (A)(B)(C) but neither $(5,7,73)$ nor $(7,5,73)$ satisfies (b), since $7.73$ equals 1 modulo 5 and $5.73$ equals 1 modulo 7. 
