Prove that a group of order 42 has a non trivial normal subgroup I know there are various solutions to this problem. We are to use Cauchy theorem. I am doing my best to understand the proof but am missing some key logic points. The parts that lose me are the following author's claims:


“... $|S| \le 6$ since if $H_1, H_2$ are any two different subgroups of G of order 7 then $H_1 \cap H_2 = (e)$
and so $\cup_{H\in S} H = 6 |S| + 1$...” 
I understand why it has to be $\leq 6$, I just don’t understand the  rationale.
And the claim that $\varphi$ is not injective:
”Now, $\varphi$ is not injective since otherwise, by Lagrange’s Theorem, $42 | n!$ for some $n \le 6$, and so $7 | n!$ for some $n \le 6$ which is obviously impossible...”
This part confuses me because other proofs I’ve read claim the opposite, that it is indeed injective. What am I missing here?


Could someone clear up those arguments for me?

Here’s Author's proof:
By Cauchy’s Theorem there are subgroups of $G$ of order $7$. 

Let $S = \{H \le G : |H| = 7\}$. Observe that $|S| \le 6$ since if $H_1, H_2$ are any two different subgroups of $G$ of order $7$ then $H_1 \cup H_2 = (e)$ and so $\cup_{H\in S} H = 6 |S| + 1$. 

Define $\varphi :G\to A(S)$ by $\varphi(x)(H)=xHx^{−1}$ for each $H \in S$. 

Note that $\varphi$ is well-defined since for each $x\in G$, $H_1, H_2 \in S$, we have that $\varphi(x)(H_1) = \varphi(x)(H_2)$ iff $xH_1x^{−1} = xH_2x^{−1}$ iff $H_1 = H_2$ and so $\varphi(x)$ is a well-defined injection from $S$ into $S$ and because $S$ is finite, $\phi (x) \in A (S)$. 

Furthermore $\varphi$ is a group homomorphism, since for $x, y \in G$ and $H \in S$ we have 
$$
\varphi (xy) (H) = xyHy^{−1}x^{−1} = \varphi (x) yHy^{−1} =
\big(\varphi (x) \circ \varphi (y)\big) (H).$$ 

Now, $\varphi$ is not injective since otherwise, by Lagrange’s Theorem, $42 | n!$ for some $n \le 6$, and so $7 | n!$ for some $n \le 6$ which is obviously impossible. 

Observe that if $\varphi (x) = iS$ for all $x\in G$ then $xHx^{−1} = H$ for all subgroups $H$ of $G$ of order $7$. Hence, in this case, every subgroup of $G$ of order $7$ is normal. 

On the other hand, if $\varphi (x)\neq iS$ for all $x\in G$ then $(e)\neq \ker(\varphi)\rhd G$ and we are done. $\blacksquare$

I tried to clean up the formatting as best I knew how. Thanks
For the first part, a group of order $7$ has $6$ nonidentity elements, and two different subgroups of order $7$ intersect only in the identity, so the number of elements accounted for by all subgroups of order $7$ is $6n+1$, where $n$ is the number of subgroups. This must be less than or equal to $42$, so $n\leq 6$.

The action of $G$ on the set of subgroups of order $7$ by conjugation induces a homomorphism into $S_n$, where $n\leq 6$. The source you found that says it is injective is mistaken, because $S_n$ for $n\leq 6$ has no element of order $7$ because its order is not divisible by $7$. If $n&gt;1$, this implies that there is a proper nontrivial normal subgroup.
|G| =42=2×3×7
The prime 7/|G|
So,there exist a subgroup of order 7.
Now, Nomber of 7-SSG=1+7k/6. For k=0,1,2,.......
                                         =1 ,for k=0
So 7-SSG is unique.So, the subgroup of order 7 is normal in G.
That's the proof.
THANKS.