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Suppose that $G$ is a group of order $60$ and $G$ has a normal subgroup $N$ of order $2$. Show that $G$ has normal subgroups of order $6$ and $10$

no of $3$-Sylow subgroups are $1$ ,$4$ or $10$ (by Sylow's third theorem)

no of $5$-Sylow subgroups are $1$ or $6$.

If I can prove no of $3$-Sylow subgroups is $1$ and no of $5$-Sylow subgroups is $1$ then both would be normal and I can multiply it with $N$ to get normal subgroups of order $6$ and $10$.

I am trying to use a cardinality argument to prove the same but I am stuck.

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    $\begingroup$ Note that if will suffice to prove that the quotient $G/H$ (where $H$ is the normal subgroup of order $2$) has a normal Sylow-$3$ and a normal Sylow-$5$. $\endgroup$ – Tobias Kildetoft Apr 11 '14 at 8:02
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    $\begingroup$ Hmm, so I cannot see a very direct argument for that actually (i.e. that a group of order $30$ has normal subgroups of orders $3$ and $5$). The best I can see is first showing that it has at least one of them and then getting the other by taking a further quotient and using that groups of order $15$ are cyclic. $\endgroup$ – Tobias Kildetoft Apr 11 '14 at 8:18
  • $\begingroup$ Yes, I think that's the way to go - what don't you like about it? $\endgroup$ – jpvee Apr 11 '14 at 9:01
  • $\begingroup$ @Tobias actually i m not getting how proving that the quotient G/H (where H is the normal subgroup of order 2) has a normal Sylow-3 and a normal Sylow-5 is enough... $\endgroup$ – Kayoken Apr 11 '14 at 9:59
  • $\begingroup$ Those subgroups then correspond to normal subgroups of the desired orders in $G$ by the correspondence theorem. $\endgroup$ – Tobias Kildetoft Apr 11 '14 at 10:54
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fact$1$: Any group of order $30$ order has normal subgroup of order $15$.

proof: If it has a normal subgroup $H$ of order $5$ and $K$ be any sylow-$3$ subgroup of $G$ then $HK$ is a subgroup of $G$ with order $15$ and since $[G:HK]=2$ then $HK$ is normal in $G$. So assume that it has no normal subgroup of order $5$ then $n_5=6\implies$ $6*(5-1)=24$ elements of order $5$. This must require that $n_3=1$ (otherwise you have more than $30$ elements in total.) So, $HK$ is again a subgroup ...

fact$2:$ Any group of order $15$ is cyclic. (you can show it in many ways including Sylow Theorems.)

Now, Let $G$ be a group of order $60$ and $N$ be a normal subgroup of order $2$.

$G/N$ is a group of order $30$ by fact$1$ it has a normal subgroup $H/N$ of order $15$ so $H$ is a normal subgroup of $G$ with order $30$.

Since $|H|=30$ it has a normal subgroup $L$ of order $15$ and $L$ is cyclic and $H$ includes $N$, since $L\cap N=1$, $H\cong L\times N\cong Z_{15}\times Z_2\cong Z_{30}$.

conclusion: Notice that $H$ includes all Sylow-$5$ subgroups since $H$ is normal in $G$ and $5$ divides $|H|$. With same reasoning, it includes all sylow-$3$ subgroups of $G$ . As $H$ has uniqe sylow $p$ subgroups we are done. From that point you can also see that $G\cong Z_{60}$ or $G\cong D_{60}$ or $G\cong Z_2\times Z_{30}$.

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