Problem 5.3.1 (Groups of order 16)

Question

Prove that $D_{16}, Z_2 	imes D_8, Z_2 	imes Q_8, Z_4 * D_8, QD_{16}$ and $M$ are nonisomorphic non-abelian groups of order $16$ (where $Z_4 * D_8$ is described in Exercise 12 Section 1, and $QD_{16}$ and $M$ are described in the exercises in Section 2.5).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}
\begin{proof} 
We compute (with Sagemath) the orders of the elements in each group:
\begin{center}
\begin{tabular}{|c|c|ccc|}
\hline
 & Description  & Order 2 & Order 4  & Order 8 \
\hline
$D_{16}$ &$\Z/(8) times _{-1} \Z/(2)$  & 9  & 2 & 4 \
 $Z_2 	imes D_8$& $Z_2 	imes D_8$ &  11&  4 & 0\
 $Z_2 	imes Q_8$&  $Z_2 	imes Q_8$& 3 &  12& 0\
 $Z_4  * D_8$ &$(Z_4 	imes D_8)/Z$ & 7 & 8 & 0\
 $QD_{16}$ &$\Z/(8) times _{3} \Z/(2)$ & 5 & 6 & 4\
 $M$ & $\Z/(8) times _{5} \Z/(2)$ & 3 & 4 & 8\
\hline
\end{tabular}
\end{center}
This is sufficient to show that these six groups are not pairwise isomorphic.

\end{proof}

Note: there are 9 isomorphism types of non abelian groups of order 16: see Keith Conrad 

https://kconrad.math.uconn.edu/blurbs/grouptheory/group16.pdf


	Description	Order 2	Order 4	Order 8

$D_{16}$	$ℤ ∕ (8) ⋊_{- 1} ℤ ∕ (2)$	9	2	4
$Z_{2} \times D_{8}$	$Z_{2} \times D_{8}$	11	4	0
$Z_{2} \times Q_{8}$	$Z_{2} \times Q_{8}$	3	12	0
$Z_{4} * D_{8}$	$(Z_{4} \times D_{8}) ∕ Z$	7	8	0
$Q D_{16}$	$ℤ ∕ (8) ⋊_{3} ℤ ∕ (2)$	5	6	4
$M$	$ℤ ∕ (8) ⋊_{5} ℤ ∕ (2)$	3	4	8

Problem 5.3.1 (Groups of order 16)

Answers

Comments

Problem 5.3.1 (Groups of order 16)

Answers

Comments

Add answer