Since $105 = 3\cdot5\cdot7$, another result implies that there must exist a normal $7$-Sylow subgroup. Does this help prove that $G$ is abelian? Also, can this result be generalized (If $|G| = p\cdot q\cdot r$, and $G$ has a normal $p$-Sylow subgroup, then $G$ is abelian (where $p<q<r$).

  • $\begingroup$ I suppose you mean "$G$ has a normal $p$-Sylow subgroup" in your last sentence? $\endgroup$ – Chris Eagle Dec 12 '12 at 11:14
  • $\begingroup$ If $p-1$ does not divide either $q$ or $r$ and $q$ does not divide $r-1$ then it is true that $G$ is abelian when $G$ has a normal Sylow p-subgroup. $\endgroup$ – peoplepower Dec 12 '12 at 11:17
  • $\begingroup$ @ChrisEagle Yes, you're right. $\endgroup$ – chubbycantorset Dec 12 '12 at 11:20
  • $\begingroup$ @peoplepower: could you elaborate on why that's true? $\endgroup$ – chubbycantorset Dec 12 '12 at 11:22
  • 1
    $\begingroup$ @chubbycantorset I was following closely the proof given by Alex Youcis here $\endgroup$ – peoplepower Dec 12 '12 at 11:26

Note that $n_5=1$ or $21$, and $n_7=1$ or $15$. Note that by order considerations they cannot both be distinct than one, so either one of them is normal. Thus, $P_5P_7\leq G$ is a subgroup of order $35$ which we shall call $K$.

Let $H$ denote the $3$- Sylow subgroup, then we have that $|HK|=|G|$ (using the fact that they intersect trivially since their orders are $3$ and $35$). Recall that if $H$ is normal in $G$ and if $|HK|=|G|$ with $H\cap K=\{e\}$, then $G\cong H\rtimes_{\varphi} K$, where $\varphi: K\rightarrow Aut(H)$. Note that since $H\cong C_3$, we have that $Aut(H)\cong C_2$. Since the order of the image of an element divides the order of the element, we must have that $\varphi$ is the trivial map. Hence there is at most one group of of order $105$ that has a normal $3$ Sylow subgroup, and we know that $C_{105}$ satisfies it, so $G$ must indeed be this group.

As mentioned in the comments below I used the fact without much mention that there is only one subgroup of order $35$. Let me prove this.

Let $G$ be a group of order $35$, then let $H_7$ denote a Sylow subgroup, note that there can only be one such group since $n_7$ divides the order of the group and since $n_7=1+7k$ for an integer number $k$. Then let $H_5$ be a sylow subgroup of $G$. In total we have $H_7$ a normal subgroup of $G$, $H_5$ and $H_7$ intersect trivially so $|H_5H_7|=|G|$ so $G$ is going to be isomorphic to $H_7\rtimes_\varphi H_5$, where $\varphi:H_5\rightarrow Aut(H_7)\cong C_6$, since the order of the iimage of an element divides the order of the element, we see that only the trivial maps exists, so we see that at most one group of order $35$ exists, and hence it must be $C_{35}$.

Note that in general you can use this idea to prove that groups of order $pq$ with $p$ not dividing $q-1$ are going to be cyclic.

  • $\begingroup$ Wait, I am reading your question again and I am a little puzzled. In the title of the question it says "Show that it is abelian if it has a $3$ Sylow normal subgroup" but then you dont say that on the body. $\endgroup$ – Daniel Montealegre Dec 12 '12 at 11:41
  • $\begingroup$ I didn't see a point in repeating the statement, and added other info instead. Also, you've shown that there is at most one group of order $105$ with a normal $3$ Sylow subgroup, but my question is: If $G$ has a $3$ Sylow subgroup, then $G$ is abelian. $\endgroup$ – chubbycantorset Dec 12 '12 at 11:50
  • $\begingroup$ Well if there is at most one group of order $105$ with a normal $3$ sylow subgroup, and since $C_{105}$ is a group of order $105$ that has a normal $3$ sylow subgroup, then we must have that $G\cong C_{105}$. $\endgroup$ – Daniel Montealegre Dec 12 '12 at 11:52
  • $\begingroup$ How in your first paragraph do you conclude that $K$ is normal? $\endgroup$ – Chris Eagle Dec 12 '12 at 14:38
  • $\begingroup$ Also, you seem to be tacitly using the fact that there's only one group of order $35$. $\endgroup$ – Chris Eagle Dec 12 '12 at 14:38

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