Show that every group of order 18 is isomorphic to a semidirect product of two abelian groups

I really don't have any clue how to start this. I think I must show that for each group of order 18, a group isomorphism to a semi direct product of two abelian groups can be found. But how do I start with this? 'every group of order 18' seems just too general for me...

Can you please help me to go on?


Hint: What can you say about the $p$-Sylow subgroups of your group $G$ for $p = 2$ and $p = 3$?

[I'm assuming you've encountered the Sylow Theorems at this point; if not, the Wikipedia article gives a nice quick summary. Herstein's Algebra gives rather more detail, as do most other Algebra books.]

Specifically: How many elements does a 3-Sylow subgroup have? The third Sylow theorem says that the number of Sylow $p$-subgroups of a group for a given prime $p$ is congruent to 1 mod $p$. How many 3-Sylow subgroups can there be?

Is either the 2-Sylow or 3-Sylow subgroup necessarily normal?

Once you get a normal subgroup $K$, the rest is pretty much downhill, since all you need is a complementary subgroup, $H$, and a homomorphism from $H$ to the automorphisms of $K$. If $K$ had, say, 9 elements, then $H$ would be of order 2, and each automorphism of $K$ would lead to a possibly different 18-element group.

How many automorphisms of an 9-element group are there? In fact, how many 9-element groups are there?

To specifically address your request ("Can you please help me to go on?"), I'll start your analysis for you:

Let $G$ be a group of order 18. We'll show that $G$ is isomorphic to one of [fill in] different groups.

Let $K$ be a 3-Sylow subgroup of $G$; the order of $K$ is [fill in]. The index of $K$ in $G$ is therefore [fill in]. The third Sylow theorem tells is that the number $n_3$ of $3$-Sylow subgroups is congruent to 1 mod 3, and the second tells us that $n_3$ divides the index of $K$ in $G$. We can therefore conclude that $n_3 =$ [fill in list of possible values].

Let $H$ be a $2$-Sylow subgroup, which necessarily has order $2$, and let $u$ be the non-identity element of $H$, which therefore satisfies $u^2 = e$.

The conjugate, $uK u^{-1}$ of $K$ is also a 3-Sylow subgroup of $G$. From this, we know that $u K u^{-1} = $[fill in possibilities here]. ...

  • $\begingroup$ Thank you for your great answer! As far as I understand there are 2 2-Sylow subgroups of G and 1 3-Sylow subgroup of G $\Rightarrow$ the 3-Sylow subgroup is normal (the 2-Sylow subgroups are not). But how do I get to the number of elements of that group? I thought about Lagrange, but as I don't know the group, I cannot calculate the index. I looked through our book and tried to find anything that helped me to get to the number of groups G with #G = 9, but unfortunately didn't find something. According your last step: As there is only one 3-Sylow subgroup, we know that $uKu^{-1} = K$? $\endgroup$
    – muffel
    Dec 2 '14 at 18:25
  • $\begingroup$ The 3-sylow subgroup is indeed normal. And since $3^2$ divides 18, the first sylow theorem -- For any prime factor $p$ (with multiplicity $n$) of the order of a finite group $G$, there exists a Sylow p-subgroup of G, of order $p^n$ -- tells you that $K$ has order $9$. So your inference in the last step is correct. At this point, every element of $G$ can be written as $hk$, where $h \in H$ and $k \in K$, and the rules for multiplying $h_1 k_1$ by $h_2 k_2$ are interesting only when $h_1 = h_2 = u$; the rule in that case gives you $\phi(u)$ in the semidirect product decomposition. $\endgroup$ Dec 2 '14 at 21:38
  • $\begingroup$ Groups of order 9: Since 9 is the square of a prime, the group $K$ is abelian. (For a proof, see proofwiki.org/wiki/Group_of_Order_Prime_Squared_is_Abelian, or just use google to search for "group of order p^2".) By the structure theorem for finitely-generated abelian groups, that means that $K = \mathbb Z/9\mathbb Z$ or $K = (\mathbb Z/3\mathbb Z) \oplus (\mathbb Z/3\mathbb Z)$. Each of these has relatively few automorphisms, so the number of possible semi-direct products is fairly small. $\endgroup$ Dec 2 '14 at 21:44
  • 1
    $\begingroup$ The cosets of $K$ form a partition of $G$. Since $u \notin K$, the cosets are $eK$ and $uK$. Things in the first coset have the form $hk$ where $h = e$; things in the second have the form $hk$ where $h = u$. :) $\endgroup$ Dec 3 '14 at 21:15
  • 1
    $\begingroup$ Now let $k_1, k_2 \in K$, and look at $(uk_1) (uk_2) = hk_3$. Claims to be proved by you: (i) $h = e$ (and more generally, if $h_1 k_1 h_2 k_2 = h_3 k_3$, then $h_3 = h_1 h_2$. I believe you'll need to use the fact that $K$ is normal.) (ii) writing $k_3 = k_1 \phi_u(k_2)$ defines a function $\phi_u$. The function $\phi_u$ is an automorphism of $K$. (iii) Your group $G$ is isomorphic to $H \rtimes_\phi K$, where $\phi_e$ is defined to be the identity, with the isomorphism being $(h, k) \mapsto hk$. $\endgroup$ Dec 3 '14 at 21:27

Alright, this is fun with Sylow and the recognition theorem for semi direct products. Sylow says that a Sylow 2-subgroup exists and that the Sylow 3-subgroup is normal. Their intersection is the identity by Cauchy's theorem, so is the semi-direct product of a cyclic 2 group acting on a Sylow 3-subgroup.


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.