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How many normal subgroups does a non-abelian group $G$ of order $21$ have other than the identity subgroup $\{e\}$ and $G$?

1) 0

2) 1

3) 3

4) 7

I think option 1 is incorrect because every group of order 1 to 59 is not simple. Hence group of order 21 has atleast one normal subgroup.

Now from the Sylow's theorem we get there has one subgroup of order 7 and one subgroup of order 3 or 7 subgroup of order 3. Since 3 and 7 are primes then any group of order 3 and 7 is cyclic. hence normal. I think I am wrong some where but I am unable to locket the point. please help me to choose the right answer.

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(It's not true that any cyclic subgroup of a group is normal. You can see this since for example the symmetric group $S_3$ has a cyclic subgroup $\{e,(12)\}$ of order 2. It is not normal because $(23)(12)(23)^{-1}=(13)$)

By Lagrange's theorem, the non-trivial proper subgroups have order 3 or 7.

As you have correctly identified, from Sylow's theorems, $G$ has a unique subgroup $N$ of order 7. It must be normal, since for any prime number $p$ the Sylow $p$-subgroups of a group form a single conjugacy class of subgroups.

Suppose (for contradiction) that it also has a normal subgroup $K$ of order 3. Then $N \cap K =\{e\}$ (By Lagrange's theorem, the order of $N \cap K$ divides 3 and 7, so is 1). Their product $NK$ is thus the whole group $G$, since it has order $\frac{|N||K|}{|N \cap K|}=\frac{3\times7}{1}=|G|$. (To see this consider the map $f:N\times K \to G, (n,k)\mapsto nk$.) So any $n \in N$ and $k \in K$ commute. (Consider an element of the form $nkn^{-1}k^{-1}$. It is in $N\cap K$, so is $e$.)

We would thus have $G$ is isomorphic to the direct product of $N$ and $K$. In particular $G$ would be abelian, which is a contradiction.

So $G$ has no normal subgroup of order 3, and by Sylow's theorems has only one of order 7. Hence option 2 is correct.

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