Hall-Senior classification of groups of order 16

This article gives specific information, namely, classification, about a family of groups, namely: groups of order 16.
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This article describes the classification of the groups of order 16 by using the ideas of Hall-Senior genus and Hall-Senior family. This is one of many mutually similar classification approaches.

Statement of classification

Abelian groups

There is a unique Hall-Senior family, called $\Gamma _{1}$ .

The nature and classification of the five abelian groups of order $p^{4}$ is the same for both the $p=2$ and odd $p$ cases; the abelian groups are classified by the set of unordered integer partitions of the number 4. This follows from the structure theorem for finitely generated abelian groups. We do not discuss the classification of abelian groups further in this article, since it is common across all classification approaches.

Partition of 4	Corresponding abelian group (in general)	Corresponding abelian group case $p=2$	GAP ID (2nd part) case $p=2$	Hall-Senior symbol	Hall-Senior number
4	cyclic group of prime-fourth order	cyclic group:Z16	1	$(4)$	5
3 + 1	direct product of cyclic group of prime-cube order and cyclic group of prime order	direct product of Z8 and Z2	5	$(31)$	4
2 + 2	direct product of cyclic group of prime-square order and cyclic group of prime-square order	direct product of Z4 and Z4	2	$(2^{2})$	3
2 + 1 + 1	direct product of cyclic group of prime-square order and elementary abelian group of prime-square order	direct product of Z4 and V4	10	$(21^{2})$	2
1 + 1 + 1 + 1	elementary abelian group of prime-fourth order	elementary abelian group:E16	14	$(1^{4})$	1

Class two groups

There is a unique Hall-Senior family, called $\Gamma _{2}$ . This family is characterized as follows: the derived subgroup is isomorphic to cyclic group:Z2, the inner automorphism group is Klein four-group. If $P$ is the whole group, then the commutator map $P/Z(P)\times P/Z(P)\to [P,P]$ is fixed as the unique alternating bilinear map possible.

Here is a classification of the groups in this family:

Group	Hall-Senior symbol	Hall-Senior number	Hall-Senior genus
direct product of D8 and Z2	$16\Gamma _{2}a_{1}$	6	$16\Gamma _{2}a$
direct product of Q8 and Z2	$16\Gamma _{2}a_{2}$	7	$16\Gamma _{2}a$
central product of D8 and Z4	$16\Gamma _{2}b$	8	$16\Gamma _{2}b$
SmallGroup(16,3)	$16\Gamma _{2}c_{1}$	9	$16\Gamma _{2}c$
nontrivial semidirect product of Z4 and Z4	$16\Gamma _{2}c_{2}$	10	$16\Gamma _{2}c$
M16	$16\Gamma _{2}d$	11	$16\Gamma _{2}d$

Class three groups

There is a unique Hall-Senior family, called $\Gamma _{3}$ . This family is characterized as follows: the derived subgroup is isomorphic to cyclic group:Z4, the inner automorphism group is dihedral group:D8, and the commutator map $P/Z(P)\times P/Z(P)\to [P,P]$ is fixed as the unique alternating bilinear map possible.

In fact, there is a unique Hall-Senior genus.

Here is a classification of the groups in this family:

Group	Hall-Senior symbol	Hall-Senior number	Hall-Senior genus
dihedral group:D16	$16\Gamma _{3}a_{1}$	12	$16\Gamma _{3}a$
semidihedral group:SD16	$16\Gamma _{3}a_{2}$	13	$16\Gamma _{3}a$
generalized quaternion group:Q16	$16\Gamma _{3}a_{3}$	14	$16\Gamma _{3}a$

Facts used

Proof of classification for nilpotency class two

Proof of the uniqueness of the Hall-Senior family

'Given: A non-abelian group $P$ of order $16$ , nilpotency class two.

To prove: If $Z(P)$ denotes the center of $P$ , then $P/Z(P)$ is isomorphic to the Klein four-group, the derived subgroup $[P,P]$ is isomorphic to cyclic group:Z2, and the alternating bilinear map $P/Z(P)\times P/Z(P)$ to $[P,P]$ is defined as follows: the image of a pair of unequal non-identity elements is the non-identity element of $[P,P]$ , and the image of equal elements and the image of a pair containing the identity element is the identity element.

Proof: Let $Z=Z(P)$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$Z$ is nontrivial.	Fact (1)	$P$ has order $2^{4}$		Fact-direct
2	$Z$ cannot have order 16.	--	$P$ is non-abelian
3	$Z$ cannot have order 8.	Facts (2), (3), (4)	$P$ is non-abelian, has order 16.		[SHOW MORE] $P/Z$ is of order 2 by Fact (4) and it is therefore cyclic by Fact (3). By Fact (2), this forces $P$ to be abelian, contradicting the assumption.
4	$Z$ has order either 2 or 4.	Fact (4)	$P$ has class two.	Steps (1),(2),(3)	[SHOW MORE] By Fact (4), the order of $Z$ must be 1,2,4,8, or 16. By the previous three steps, $Z$ has order either 2 or 4.
5	If $Z$ has order 2, then $[P,P]$ has order 2.		$P$ is non-abelian and has class two		[SHOW MORE] If $Z$ has order 2, then, since $[P,P]\leq Z$ because $P$ has class two, and further, $[P,P]$ must equal $Z$ . This forces $[P,P]$ to have class two. On the other hand, if $Z$ has order 4, the
6	If $Z$ has order 2, then we get a contradiction				PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
7	If $Z$ has order 4, then $P/Z$ is isomorphic to a Klein four-group, $[P,P]$ has order two, and the commutator map $P/Z\times P/Z\to [P,P]$ is as described.	Facts (1), (4), (5), (6)			[SHOW MORE] By Fact (4), $P/Z$ has order 4. By Fact (1), $P/Z$ is not cyclic, so by Fact (5), it must be the Klein four-group (the elementary abelian group). Further, the commutator map $P/Z\times P/Z\to [P,P]$ is an alternating bilinear map whose image generates $[P,P]$ . Hence, it gives a surjective homomorphism $\bigwedge ^{2}(P/Z)\to [P,P]$ . $\bigwedge ^{2}(P/Z)$ is isomorphic to cyclic group:Z2, and the only nontrivial map from it is an isomorphism to $[P,P]$ . Thus, $[P,P]$ is cyclic of order two.