Showing $\left\lvert G\right\rvert=105$ is not simple. Proof verification

Question

I want to know if my proof showing that a group, $G$, where $\left\lvert G\right\rvert=105$ cannot be simple, is correct.

$\left\lvert G\right\rvert=105=3\times5\times7$ gives us:

$n_3\mid35,\,n_3\equiv 1 \pmod 3,\quad n_5 \mid21,\, n_5\equiv 1 \pmod 5,\quad n_7\mid15,\, n_7\equiv 1 \pmod 7$

We get: $n_3=1$ or $7$, $n_5=1$ or $21$ and $n_7 = 1$ or $15$.

Now if $n_5 =1$ we are done, and if $n_5 = 21$ we know that since these sylow $5$-subgroups are cyclic, they intersect trivially, and hence contribute $21\times 4 = 84$ non-trivial elements, and hence we need to fill the other $21$ elements(including the trivial identity), which means $n_3=7$ or $n_7=3$. $n_7$ can't equal $3$, so $n_3=7$. But if $n_3=7$ and $n_5=21$, then there is no sylow $7$-subgroup, which is a contradiction of Sylow's theorem. Hence $n_5=1$ and therefore we always have a normal sylow $5$-subgroup in $G$, and $G$ is not simple.

Answer 1

I don't follow it starting with the part that says "which means $n_3=7$ or $n_7=3$". I don't see why $n_7$ would ever be $3$. And I do see that $n_3$ needs to be $7$, but it's not a logical consequence of what you said right before this part.

At the same point you deduced "nonsimple$\implies n_5=21$", you can also deduce "nonsimple$\implies n_3=7$" and "nonsimple$\implies n_7=15$". So the same method that gave you $84$ elements of order $5$ gives you $14$ elements of order $3$ and $90$ elements of order $7$. It's all too many elements.

Answer 2

This is another way:

Suppose $n_3=7$. Since the number of Sylow-$p$ subgroups is equal to the index of the normalizer of a Sylow-$p$ subgroup, so if $N$ is the normalizer of Sylow-$3$ subgroup, then $[G\colon N]=7$.

$G$ permutes the $7$ (left) cosets of $N$ by left multiplication. Hence we get a homomorphism $$G\rightarrow S_7$$ Now $|N|=3.5=15$, $N$ must be cyclic. If the above homomorphism has kernel equal to identity, then it will be injective, and hence $N$ will also be (isomorphic to) a subgroup of order $15$ in $S_7$.

One can see that there is no element of order $15$ is $S_7$.

Further, the kernel of above homomorphism can not be whole $G$, since if you take $x\in G$ outside $N$, then $x$ will take the coset $N$ to $xN$ which is different from $N$, i.e. $x$ can not be in kernel of the homomorphism.

Hence above homomorphism has non-trivial kernel, $G$ is not simple.