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...
Can you please help me to go on?
 A: 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].  ...
A: 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. 
