# Why are all groups of order 153 abelian?

$153 = 3^2 \cdot 17$ so lets assume there are $s_3$ $3$-Sylow-Subgroups and $s_{17}$ $17$-Sylow-Subgroups. We know that $s_3 \mid 153$ so $s_3 \in \{1,3,9,17,51,153\}$ and $s_{17} \in \{1,17,51\}$. Due to the rule that the number of $p$-Sylow-Subgroups $s$ must satisfy

$$1 + pk = s$$

for some $k \in \mathbb{N}_0$. That leaves $1 = s_3 = s_{17}$. Is that correct? Now those two Subgroups $H_3, H_{17}$ are cyclic and abelain. How do I draw a conclusion about all possible groups of order 153 with this information?

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Sylow theory says that it has normal (=unique) Sylow p-subgroups $N$ and $S$ of order 9 and 17 respectively. Since $G$ is clearly a direct sum of these two subgroups, it suffices to show that the subgroups are Abelian.
$S$ is cyclic, hence abelian. You should be able to tell us why $N$ (the Sylow of order 9) is also abelian.
Hint: The center of $N$ is nontrivial, so it could have order 3 or 9. If it's order 9 then we're done (because it's abelian!) Thus, you just have to show why the center can't have order 3.
This is a classic exercise that would be relevant for this last part. One version of it goes "If $Z$ is a subgroup of the center of $G$ and $G/Z$ is cyclic, then $G$ is abelian."
@tomasz They certainly have trivial intersection, and so their product is a subgroup of $G$ of order $9*17$, therefore there are not many options! – rschwieb Aug 24 '12 at 14:31
@tomasz I think the concerns you have are dealt with when proving the basic formula that works in finite groups: $|NH|=|N|\cdot|H|/|N\cap H|$ – rschwieb Aug 24 '12 at 16:24
@tomasz They're both normal, $N\cap H=\{1\}$ and $NH=G$... I'm now aware of any additional criterion necessary for identifying it as an internal direct sum... – rschwieb Aug 24 '12 at 18:34