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.
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$\begingroup$ Its not homework. I got this problem from gallians book $\endgroup$ – mathscrazy Apr 21 '13 at 14:55
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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.
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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
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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 $r = 7$.
answered Apr 21 '13 at 18:15
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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.
answered Apr 21 '13 at 15:15
