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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.

Sourav Ghosh
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Sayan rakshit
  • 177

3 Answers3

12

(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.

Matt Gallagher
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2

Without Sylow. A non-abelian group of order $21=3\cdot7$ is centerless. In fact, if $|Z(G)|=3$ (respectively, $7$) then all the noncentral elements have the centralizer of order $3$ (respectively, $7$), and hence the class equation yields $21=3+7k$ (respectively, $21=7+3k$): contradiction, because $7\nmid 3$ (respectively, $3\nmid 7$).

Now, recall that a subgroup is normal if and only if it is the union of conjugacy classes (see e.g. here). The noncentral conjugacy classes of $G$ have size $3$ or $7$ (orbit-stabilizer), so $P\unlhd G$ if and only if: $$|P|=1+3k+7l \tag1$$ for some nonnegative integers $k$ and $l$, with $|P|=3$ or $7$. Now, $(1)$ has no solutions $(k,l)$ for $|P|=3$, whilst it has the only solution $(k,l)=(2,0)$ for $|P|=7$. So, your $G$ has one normal subgroup, only (of order $1+3\cdot2=7$, for the records). The correct option is then 2.

citadel
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Let $G$ be a non abelian group of order $21$.

$|G|=21=3\cdot 7$

Then by Sylow's theorem, number of sylow $7$ -subgroup of $G$ , $$n_7:=(1+7k) |3$$ for $k\in \Bbb{N}\cup\{0\}$

Clearly $n_7=1$ and hence subgroup of order $7$ is normal.

$n_3= 1,7$

If $n_3=1$ i.e subgroup of order $3$ is normal then $G$ would be abelian as $G$ would be direct product of two abelian ( infact cyclic as both are group of prime orders) groups.

In other words $G\cong \Bbb{Z}_3×\Bbb{Z_7}$ which contradict $G$ is non abelian.

Hence $G$ has only one normal subgroup except two trival subgroups $\{e\}, G$

Sourav Ghosh
  • 12,997