Anyone who has had to prepare for an algebra qualifying exam is familiar with the "Classify groups of order $X$" question.

To illustrate my general question, which I postpone until the end, consider the following simple example in which I classify groups $G$ of order $3 \cdot 7$. Let $H$ and $K$ be the $7$- and $3$-Sylow subgroups, respectively. By Sylow's theorems, we find easily that $H$ is normal and $K$ is either normal or one of seven conjugate copies. Also $H \cong \mathbf{Z}_7$ and $K \cong \mathbf{Z}_3$. Let $x$ be a generator of $\mathbf{Z}_3$ and let $y$ be a generator of $\mathbf{Z}_7$, both viewed multiplicatively. Now $G$ is a semidirect product of $H$ and $K$, hence the possible structures of $G$ are determined by the possible group homomorphisms $$ \mathbf{Z}_3 \to \mathrm{Aut}(\mathbf{Z}_7) \cong \mathbf{Z}_6. $$ Such a group homomorphism is determined by the image of $x$; since the order of this image must divide the order of $x$, we see $x$ is either sent to the identity automorphism $\mathbf{1}$ or an automorphism of order three.

We find that a generator of $\mathbf{Z}_6$ is the automorphism $\alpha \colon y \mapsto y^3$. Therefore, there are three possible group homomorphisms, determined by sending $x$ to $\mathbf{1}$, to $\alpha^2 \colon y \mapsto y^2$, or to $\alpha^4 \colon y \mapsto y^4$. It follows there are at most three possible groups of order $21$, generated by $x$ and $y$ and subject to the relations $x^3 = x^7 = 1$ as well as one of the following commutativity relations: $$xy = yx, \;\;\;\;\; xy = y^2 x, \;\;\;\;\; xy = y^4 x. $$ All such groups exist by employing the abstract construction of the semidirect product.

What follows is always the most subtle part of the analysis.

Which of these groups are duplicates?

The first is the case $G \cong \mathbf{Z}_3 \times \mathbf{Z}_7$ which is clearly distinct from the remaining two. Let the second group be denoted $G_2$ and the third $G_4$. If $G_2$ were isomorphic to $G_4$, then there would have to exist $X, Y \in G_4$ of orders three and seven, respectively, and satisfying $XY = Y^2 X$ (or $X, Y \in G_2$ satisfying $XY = Y^4 X$). And it is easy to see that, in fact, this condition is sufficient for $x \mapsto X$, $y \mapsto Y$ to determine an isomorphism $G_2 \cong G_4$. Note that both $G_2$ and $G_4$ have seven $3$-Sylow subgroups (otherwise they would be Cartesian products). So there are $14$ candidates for $X$ and $6$ candidates for $Y$.

Morally, at least in my opinion, these groups should be isomorphic, because the only difference in their definition occurs when we chose between the two generators $\alpha^2$ and $\alpha^4$ of the cyclic subgroup $\mathbf{Z}_3 \subset \mathrm{Aut}(\mathbf{Z}_7) \cong \mathbf{Z}_6$, and these generators are 'essentially the same'.

This is indeed the case, but the proof feels 'lucky'. One finds by calculation that no map of the form $X = x$ and $Y = y^k$ satisfies $XY = Y^2 X \in G_4$. But this is satisfied by taking $X = x^2$ and $Y = y$: $$X Y = x^2 y = x y^4 x = y^{16} x^2 = y^2 x^2 = Y^2 X \in G_4. $$ We conclude there are two groups of order $21$ up to isomorphism.

To give an example of how this problem becomes more complex, if instead one were computing groups of order $3 \cdot 7 \cdot 13$, then one must determine group homomorphisms $$\mathbf{Z}_3 \to \mathrm{Aut}(\mathbf{Z}_7 \times \mathbf{Z}_{13}) \cong \mathbf{Z}_6 \times \mathbf{Z}_{12}. $$ (Don't forget that the automorphisms of the direct product is the direct product of the automorphisms when the orders of the groups are coprime!) If $\alpha$ generates $\mathbf{Z}_6$ and $\beta$ generates $\mathbf{Z}_{12}$, then there are nine possible semidirect structures, corresponding to $x$ being sent to to any of the following pairs: $$(\mathbf{1}, \mathbf{1}), \;\;\;\;\; (\alpha^2, \mathbf{1}), \\ (\alpha^4, \mathbf{1}), \;\;\;\;\; (\mathbf{1}, \beta^4), \\ (\mathbf{1}, \beta^8), \;\;\;\;\; (\alpha^2, \beta^4), \\ (\alpha^4, \beta^4), \;\;\;\;\; (\alpha^4, \beta^8), \;\;\;\;\; (\alpha^2, \beta^8). $$ Which of these are isomorphic?

Hopefully at this point my general question is clear. First, in words:

In considering semidirect products $G \cong H \rtimes K$ is there a (natural?) proof that shows the choice of generator(s) of $\mathrm{Aut}(H)$ affects the resulting group only up to the choice of 'non-equivalent' generators?

Here is a precise phrasing for which I would be thrilled to receive an answer:

Question: Prove or disprove. Let $p$ and $q$ be primes such that $q$ divides $p-1$. Consider semidirect products $G_\rho = \mathbf{Z}_p \rtimes_\rho \mathbf{Z}_q$ determined by group homomorphisms $$ \rho \colon \mathbf{Z}_q \to \mathrm{Aut}(\mathbf{Z}_p) \cong \mathbf{Z}_{p-1}. $$ Let $x$ multiplicatively generate $\mathbf{Z}_q$ and let $\alpha$ multiplicatively generate $\mathbf{Z}_{p-1}$. Setting $n = (p-1)/q$ the generators for $\mathbf{Z}_q \subset \mathbf{Z}_{p-1}$ are $\alpha^{nk}$ where $k = 1, \dots, q-1$. Let $\rho_k$ denote the group homomorphism determined by $x \mapsto \alpha^{nk}$. Then $G_{\rho_k} \cong G_{\rho_\ell}$ for all $1 \leq k, \ell \leq q-1$.


From Dummit and Foote, Chapter 5, page 184:

Assume that $K$ is a cyclic group, $H$ is an arbitrary group and $\phi_1(K)$ and $\phi_2(K)$ are conjugate subgroups of $\operatorname{Aut}(H)$. If $K$ is inifite assume $\phi_1$ and $\phi_2$ are injective. Prove by constructing an explicit isomorphism that $H \rtimes_{\phi_1} K \cong H \rtimes_{\phi_2} K$ (in particular, if the subgroups $\phi_1(K)$ and $\phi_2(K)$ are equal in $\operatorname{Aut}(H)$, then the resulting semidirect products are isomorphic).

It even goes on to give a hint. (I strongly recommend spending a bit of time trying to do it yourself before reading the hint- it's a real spoiler.)

Suppose $\sigma \phi_1(K) \sigma^{-1} = \phi_2(K)$ so that for some $a \in \mathbb{Z}$ we have $\sigma \phi_1(k)\sigma^{-1} = \phi_3(k)^a$ for all $k \in K$. Show that the map $\psi:H \rtimes_{\phi_1} K \to H \rtimes_{\phi_2} K $ defined by $\psi(h,k)=(\sigma(h),k^a)$ is a homomorphism. Show $\psi$ is bijective by constructing a 2-sided inverse.

There's actually a lemma worth knowing here:

Lemma: Let $a,m,n \in \mathbb{Z}$ such that $m$ divides $n$ and $\gcd(a,m)=1$. Then there exists $\overline{a} \in \mathbb{Z}$ such that $\overline{a} \equiv a \mod m$ and $\gcd(\overline{a},n)=1$.

There's a full explanation here (this is where I first learned the lemma), but I welcome you to ask for hints rather than peak at the spoiler or answer.

  • $\begingroup$ Well this answer is everything I dreamed it would be, and more. Thanks so much for the reference! +1 $\endgroup$ – Doug Jul 28 '16 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.