# Finding a group isomorphism for every group of order 18 and the semidirect product of abelian groups

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...

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?

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]. ...

• 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 $uKu^{-1} = K$? Dec 2 '14 at 18:25
• 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. Dec 2 '14 at 21:38
• 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. Dec 2 '14 at 21:44
• 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 $h = u$. :) Dec 3 '14 at 21:15
• 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 $(h, k) \mapsto hk$. 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.