# Constructing groups of order 16 from knowledge of their subgroups

So I am rather curious about known techniques for constructing groups from knowledge of their subgroups. For instance, if I told you that a particular order 16 group had 3 order 8 subgroups, all isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_4$, it is possible to immediate conclude that the group is $2$-generated because it is an order 16 group with 3 order 8 subgroups. Moreover, it turns out that there are exactly two possibilities, but I only know this because I have analyzed all 14 groups of order 16. The possibilities are the abelian group $\mathbb{Z}_4\times\mathbb{Z}_4$ and a non-abelian semidirect product, $\mathbb{Z}_4\rtimes_\varphi\mathbb{Z}_4$, where $\varphi$ is surjective (onto $\operatorname{Aut}(\mathbb{Z}_4)\simeq\mathbb{Z}_2$).

My admittedly somewhat general question is this: what methods would sufficient to construct all 14 groups of order 16 from just knowledge of the groups of order 1, 2, 4, and 8?

I know they can all be represented as semi-direct products, but the same group can often be represented in quite a few different ways as semidirect products, which makes determining which groups are which (up to isomorphism) tedious. From analyzing the 14 different groups, most of them are readily distinguishable from each other via much more readily noticed properties (especially from a constructive perspective). For instance, it's easy to distinguish the two possibilities mentioned previously by looking at whether or not elements commute (curiously, they have identical subgroup lattices, sans that the entire group is not isomorphic). Likewise, distinguishing the modular group $M_{16}$ from the previous two is trivial if you have access to their subgroup lattices (the one for $M_{16}$ is clearly different from the other two, moreover, it has a cyclic subgroup of order 8, which the other two do not).

A closely related question: If I gave you an operation table for a 'mystery group' of order 16, what things would most quickly (quickly in terms of ease to compute) allow you to identify its isomorphism class? (I can usually get it down to at most two possibilities very quickly, from just knowledge of subgroups, but at some point, that alone may not be enough, as evidenced with the above example)

• Maybe you already know this, but the number of order-8 subgroups is encoded in the quotient of $G$ by its Frattini subgroup $\Phi(G)$ (= the intersection of all maximal subgroups of $G$). In a $p$-group, every maximal subgroup has index $p$, and the quotient $G/\Phi(G)$ is elementary abelian. So the quotients $G/M$ for $M$ maximal are exactly the order-$p$ quotients of $G/\Phi(G)$. Since $G/\Phi(G) \cong (\mathbb Z/p\mathbb Z)^r$, there are $p^r - 1$ such quotients. Thus your condition of having 3 order-8 subgroups is equivalent to having $G/\Phi(G) \cong (\mathbb Z/2\mathbb Z)^2$. – Ravi Fernando Aug 21 '16 at 21:48
• As far as identifying a mystery group, it happens amusingly often that a small group is determined uniquely by the list of orders of its elements, which seems like one of the crudest possible invariants of a group, next to its order. There certainly are non-isomorphic groups with the same lists of orders, e.g. $(\mathbb Z/4\mathbb Z)^2$ and $\mathbb Z/2\mathbb Z \times Q_8$. But those two might be the only "counterexample" of order 16, and they're distinguished by the even cruder property of abelianness. – Ravi Fernando Aug 21 '16 at 21:55
• In a more theoretical direction, the computational complexity of the group isomorphism problem (i.e. determine whether two given groups are isomorphic) has been studied fairly extensively. I'm not an expert on this (and in particular I'm not sure exactly how sensitive the problem is to presenting the group with generators and relations vs. via a multiplication table), but see e.g. cs.stackexchange.com/questions/35764/…. – Ravi Fernando Aug 21 '16 at 22:17
• @RaviFernando: More precisely, there are 9 different 'order counts' exhibited by order 16 groups, most of which uniquely identify a group, however 3 of them do not, with 2 of those 3 belonging to 3 different groups each, and the remaining order count belonging to just 2 groups. These order counts, in numbers of elements of orders, 16,8,4,2,1, are $[8,4,2,1,1],[0,8,4,3,1],[0,8,4,3,1],[0,4,10,1,1],[0,4,6,5,1],[0,4,2,9,1],[0,0,12,3,1],[0,0,12,3,1],[0,0,12,3,1],[0,0,8,7,1],[0,0,8,7,1],[0,0,8,7,1],[0,0,4,11,1],[0,0,0,15,1]$ where repeated listing indicates non-isomorphic group w/ same order counts. – Justin Benfield Aug 21 '16 at 22:47
• @RaviFernando: I am interested in easy ways to rule out plausible order counts, such as $[0,12,2,1,1]$ which is impossible, but not obviously so. (It's easy to prove that such a group, if it existed must be non-abelian, have all 3 order 8 subgroups isomorphic to cycle group of order 8, and that said subgroups all share common cyclic subgroup of order 4, but it's not so obvious that such a group is actually impossible. The analogous group of order 8 is possible, namely, the quaternion group, $Q_8$.) – Justin Benfield Aug 21 '16 at 22:50

First some comments. There are different ways of determining groups of order $16$; first perhaps given by Burnside in his book. In my opinion, the construction by Burnside is better than all (including my way, which I will describe shortly), because Burnside first determines all $2$-groups with a specific property and then concludes for a special case, when order is $16$.

My way of description of groups of order $16$ is bases on some specific subgroups with a strong connection between them. We will classify only non-abelian groups.

Fact 1. Every group of order $p^4$ contains an abelian subgroup of index $p$.

Fact 2. If a non-abelian $p$-group contains an abelian subgroup of index $p$ then $|G|=p.|Z(G)|.|G'|$.

Thus, when $|G|=2^4$, from the two facts, we get $2^4=2.|Z(G)|.|G'|$. Since $Z(G)$ and $G'$ are non-trivial, it follows that $$\mbox{ either } |Z(G)|=2, |G'|=4 \mbox{ or } |Z(G)|=4, |G'|=2.$$ Since the center of a $p$-group intersects normal subgroup non-trivially, it follows that according to above two cases we have $$\mbox{ either } Z(G)<G' \mbox{ or } G' <Z(G).$$

We will start classification on this information with case by case computation. This simplifies our job in searching non-isomorphic groups of order $16$. I will elaborate only one case, since it involves much computations.

Now first make two cases:

Case 1. $|Z(G)|=2$, $|G'|=4$. Case 2. $|Z(G)|=4$ and $|G'|=2$.

(Note that the groups obtained in case 1 can not be isomorphic to groups obtained in case 2, can you see why? This justifies my statement about simplification of isomorphism problem.)

In case 1, since $|G'|=4$, we have $G'=Z_2\times Z_2$ or $G'=Z_4$. Again make two sub-cases.

Case 1.1 $Z(G)=Z_2$, $G'=Z_2\times Z_2$.

By Fact 1, there is abelian subgroup $A$ of index $2$, i.e. order $8$. Then $G/A$ is cyclic (abelian), hence $G'<A$. Since $G'=Z_2\times Z_2$, the abelian group $A$ must be of the form $Z_2\times Z_2\times Z_2$ or $Z_4\times Z_2$.

Case 1.1.1 $Z(G)=Z_2$, $G'= Z_2\times Z_2$ and $A=Z_2\times Z_2\times Z_2$.

Let $Z(G)=\langle x\rangle$, $G'=\langle x,y\rangle$ and $A=\langle x,y,z\rangle$. Take $w\in G\setminus A$. Then $G=\langle x,y,z,w\rangle$. Now we know relations between $x,y,z$ (since these are members of an elementary abelian group). Just we have to search relations of $w$ with $x,y,z$.

Now $A=\langle x,y,z\rangle$ is normal in $G$, which means $w$ should normalize this subgroup. So we have to determine $wxw^{-1}$, $wyw^{-1}$ and $wzw^{-1}$.

• $wxw^{-1}$ must be $x$, because $x\in Z(G)$.

• Since $G'=\langle x,y\rangle$, $G'$ has three subgroups of order $2$, namely $\langle x\rangle$, $\langle y\rangle$ and $\langle xy\rangle$. Since $\langle x\rangle$ is central subgroup, $w$ should permute other two subgroups by conjugation. Now if $w\langle y\rangle w^{-1}=\langle y\rangle$ then this means $\langle y\rangle$ is normal in $G$ (because $\langle y\rangle$ is already normal in abelian subgroup $\langle x,y,z\rangle$ and is also normalized by $w$, hence by $G$). Since normal subgroup of order $2$ must be central, we have $y\in Z(G)$ a contradiction. Thus, $w\langle y\rangle w^{-1}$ must be equal to $\langle xy\rangle$. You can easily show that $wyw^{-1}=xy$.

• Since $A=G'\cup zG'$, and $wG'w^{-1}=G$', hence $w(zG')w^{-1}=zG'$. Let $wzw^{-1}=zx^iy^j$. Again conjugating by $w$ we get

$$w^2zw^{-2} = (wzw^{-1}) (wxw^{-1})^i (wyw^{-1})^j = (zx^iy^j)(x)^i(xy)^j= zx^{2i}x^jy^{2j}.$$ Since $[G\colon A]=2$, so $w^2\in A$ hence $w^2$ commutes with all elements of $A$. So $w^2zw^{-2}$ should be $z$. On the other hand in last term of above equation, since $x^2=1=y^2$, we get last term to be $zx^j$. Thus we get $$z=zx^j \mbox{ i.e. } j=0.$$ Therefore $wzw^{-1}=zx^i$ where $i=0$ or $1$. If $wzw^{-1}=z$ then $\langle z\rangle$ will be a normal subgroup of order $2$, hence $z$ will go in center, a contradiction. Hence we must have $wzw^{-1}=zx$. So far we have obtained is the following:

$$wxw^{-1}=x, wyw^{-1}=xy, wzw^{-1}=xz.$$

What about $w^2$? Since $w^2\in A$ so $w^2$ commutes with all elements of $A$. Obviously $w^2$ commutes with $w$. Hence $w^2$ must be in the center which is $Z(G)=\{1,x\}$.

Thus in Case 1.1.1, we get at most two groups of order $16$ according to $w^2=1$ or $w^2=x$. Are these two groups isomorphic? (i.e. is the group when $w^2=1$ isomorphic to group when $w^2=x$?)

Yes! Because, when $w^2=x$, then consider $w_1=wy$ instead of $w$. Then $$w_1^2=(wy)(wy)=wyw^{-1}.w^2y=xy.x.y=1.$$ Hence instead of $w$, we may take $w_1=wx$, which is still outside $A=\langle x,y,z\rangle$ and $w_1^2=1$. Thus groups in both cases are isomorphic. In other words, we say

$$\mbox{ if Z(G)=Z_2, G'=Z_2\times Z_2 and A=Z_2\times Z_2\times Z_2 then there is unique group of order 16 with this data.}$$

• This is, roughly speaking, the same method as can be found in: buzzard.ups.edu/courses/2012spring/projects/… which is an answer to my question, but I am really looking for something more concise/uses more theory to simply or outright avoid involved computations. I am also interested in ways to distinguish the 14 groups from each other (this is useful when dealing with them as subgroups of a larger group). Which these types of methods are inefficient for doing. – Justin Benfield Aug 22 '16 at 4:53
• @Justin: He is just considering $|Z(G)|$ and its isomorphism types; but here, I am also considering $|G'|$ and presence of abelian subgroup of index $p$, which makes computations simpler in the sense, we can easily say which groups are isomorphic or not isomorphic. – p Groups Aug 22 '16 at 5:06