8
$\begingroup$

I have to prove that a group of order 105 contains a subgroup of order 35. Could anybody tell me how to prove this?

Thanks.

$\endgroup$
  • $\begingroup$ Its not homework. I got this problem from gallians book $\endgroup$ – mathscrazy Apr 21 '13 at 14:55
7
$\begingroup$

You need to use Sylow theorem properly. Since $105 = 3 \cdot5\cdot 7$, the number $1 + 7k$ of 7-Sylow subgroups is 1 or 15. If there are 15 such subgroups then we get $15 \cdot 6 = 90$ elements of order 7. This makes it impossible for the number $1 + 5k$ of 5-Sylow subgroups to be 6 or more, since that would require at least $6\cdot 4 = 24$ elements of order 5, making the group too big. Thus there is a unique subgroup $G_5$ of order 5, and it is normal. For any 7-Sylow subgroup $G_7$ we thus have that $G_5G_7$ is a subgroup of order 35, and its order is 35 because $G_5 \cap G_7 = 1$. (In general, the intersection of a $p$-Sylow subgroup and a $q$-Sylow subgroup for $p \neq q$ must be trivial.) On the other hand, if there a unique subgroup $G_7$ of order 7, then it is normal. For any 5-Sylow subgroup $G_5$ we thus have that $G_5G_7$ is a subgroup of order 35 for the same reasons as in the previous paragraph but with the roles of 5 and 7 exchanged.

$\endgroup$
  • 1
    $\begingroup$ Note that it is true that every group of order $105$ will have a normal Sylow $7$-subgroup (because the subgroup of order $35$ is normal and contains a unique subgroup of order $7$). See my answer $\endgroup$ – Mikko Korhonen Apr 21 '13 at 18:15
5
$\begingroup$

If $G$ is a group of order $pqr$, where $p < q < r$ are primes, then $G$ contains a normal subgroup $R$ of order $r$ (for hints on how to prove this, see this question). Since $G$ also contains a subgroup $Q$ of order $q$, the subgroup $QR$ has order $qr$. Your problem is the case $p = 3$, $q = 5$ and $r = 7$.

$\endgroup$
  • $\begingroup$ Thanks these results are helpful for me. $\endgroup$ – mathscrazy Apr 22 '13 at 15:02
4
$\begingroup$

A group $G$ operates on the set of its $p$-Sylow subgroups by conjugation. If $G_7$ is any $7$-Sylow group, it operates on the set of $5$-Sylow groups, thus partitioning the set of $5$-Sylow groups into disjoint orbits of size either $7$ or $1$ each. Unless the number of $5$-Sylow groups is a multiple of $7$, there must exist at least one orbit of legth $1$, i.e. a $5$-Sylow group $G_5$ that is fixed by our $G_7$, i.e. $G_7$ normalizes $G_5$. But then $\langle G_5\cup G_7\rangle = G_7G_5$ has order $35$ as desired.

Similarly, unless the number of $7$-Sylow groups is a multiple of $5$, there must be a $7$-Sylow group $G_7$ that is fixed under conjugation (i.e. normalized) by a $G_5$ we picked, and again $\langle G_5\cup G_7\rangle = G_5G_7$ has order $35$.

We are left with the case that both the number of $7$-Sylow groups is a multiple of $5$ and the number of $5$-Sylow groups is a multiple of $7$. As the number of $5$-Sylows is also $\equiv 1\pmod 5$ and the number of $7$-Sylows is $\equiv 1\pmod7$, we conclude that there are at least $21$ $5$-Sylows and at least $15$ $7$-Slows, leading to at least $21\cdot 4=84$ elements of order $5$, at least $15\cdot 6=90$ elements of order $7$, which already gives us $174$ elements - contradiction.

$\endgroup$
  • $\begingroup$ Could you please explain about orbit? $\endgroup$ – mathscrazy Apr 21 '13 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.