Question

Classify all groups of order 21 up to isomorphism.

Short answer

There are two groups of order 21, ${\mathrm{\mathbb{Z}}}_{21}$ and .

Step-by-step solution

Referring to Corollary 9.16, and Theorem 9.17 (Third Sylow Theorem)

Corollary 9.16

Let G be a finite group and K be a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup.

Theorem 9.17 (Third Sylow Theorem)

The number of Sylow p-subgroups of finite group G divides $\left|G\right|$ and is of the form 1+pk for some non-negative integer k.

Classifying all groups of order 21 up to isomorphism

Let’s consider any nonabelian group of order 21.

Then, the order of G can be written as:

$\begin{array}{rcl}n& =& 21\\ & =& 3·7\end{array}$

Therefore, according to Theorem 9.17 and Corollary 9.16, any group of order 21 has a normal Sylow 7-subgroup.

So, .

Let $b\in G$be any element of order 3.

Because it is normal, it must have $b^{-1}ab=a^k$for some $k\in {\mathrm{\mathbb{Z}}}_7$.

As we have considered, G is nonabelian, so $k\ne 1$.

Since b is an element of order 3, so $b^3=e$.

Therefore, find a as:

$$\begin{gathered} a=b^{-3}a{b}^3 \\ a=a^{i^3} \end{gathered}$$

This implies, $k^3\equiv 1\left(mod\right)7$.

From this, we get three values of k, which are 1, 2, and 4.

Since $k\ne 1$, we consider the other two values.

By considering these two values, we get 2 possible combinations as:

As given, we have to consider up to isomorphism.

Isomorphism $\varphi :G_1\to G_2$is given as:

$$\begin{gathered} \varphi \left(a\right)=c \\ \varphi \left(b\right)=d^2 \end{gathered}$$

So, the nonabelian group of order up to isomorphism can be given as:

Now, if G is abelian, it must be cyclic too, and it must be isomorphic to ${\mathrm{\mathbb{Z}}}_{21}$.

Hence, there are two groups of order 21, ${\mathrm{\mathbb{Z}}}_{21}$and .