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I am trying to prove that a group of order $105$ has a subgroup of order $21$. I know it can be done using Sylow theorems, I was just wondering if the proof below could be another way of doing that.

$3$ is prime and $3$ divides $105$, so $G$ has an element of order $3$, say $x$ (By Cauchy's). Similarly, let $y$ be an element of order $7$. Then $xy$ is in $G$, and since $\gcd(3,7)=1$ order of $xy$ is $21$. Then the group $\langle xy \rangle$ generated by $xy$ has an order $21$ and is a subgroup of $G$.

I am a little concerned about my proof, since it implies that group of order $21$ is cyclic $\Rightarrow$ abelian. Is there a problem with that? I know if $p$ doesn't divide $q-1$ then it is true, but in this case $3$ divides $7-1$.

Thanks,

Srivatsan
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Jennie
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    You haven't explained why $xy$ has order $21$, and in fact it need not. – Chris Eagle Dec 27 '11 at 21:56
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    I think you're assuming that $x$ and $y$ commute. How else can you be confident of the order of $xy$? – Dylan Moreland Dec 27 '11 at 21:59
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    Note that your proof would imply that $S_3$ is cyclic. Indeed $S_3$ has a permutation $x$ of order $2$, and another one $y$ of order $3$....The mistake was explained above ... – N. S. Dec 27 '11 at 22:03
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    OK, I got it :) I don't know if x and y commute, so order of xy need not be 21. Thanks a lot. – Jennie Dec 27 '11 at 22:04
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    A finite group whose order $n$ is not divisible by the square of any prime (or more generally has no chief factor whose order is divisible by the square of any prime) has subgroups of order $d$ for every divisor $d$ of $n$, but those subgroups don't have to be cyclic (though in the square-free case they are all built from at most two cyclic subgroups, so you might want to look more carefully at the possible groups of order 21 to see the full story isn't much more complicated). – Jack Schmidt Dec 28 '11 at 04:21

2 Answers2

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Hints for an elementary approach:

Step #1: Show that if $P_3$ is a Sylow 3-subgroup, then $5\mid |N_G(P_3)|$. Thus $N_G(P_3)$ has a subgroup $H$ of order 15. A standard exercise shows that $H$ is cyclic.

Step #2: The number of Sylow 5-subgroups is either 1 or 21. By Step #1 at least one, hence all, of the Sylow 5-subgroups have a normalizer of size at least ??? Therefore the number of Sylow 5-subgroups is ???

Step #3: Identify all the automorphisms of $C_5$ of order that is a factor of seven. Conclude that if a $P_5$ is normalized by an element of order 7, then there exist elements of orders 5 and 7 that commute.

Step #4: Using steps #2 and #3 conclude that the normalizer of at least one, hence all, Sylow 7-subgroups is at least 35. Show that the group $G$ has a normal Sylow 7-subgroup $P_7$.

Step #5: Show that the quotient group $G/P_7$ has a subgroup of order 3. Apply the correspondence theorem (as in Step #1).

Nicky Hekster
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Jyrki Lahtonen
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You are explicitly asking for a Sylow-free proof of the claim. This doesn't provide a full answer, but it makes an effort to go as farthest as possible (at the best of my ability) without Sylow.

If $G$ is abelian, then $xy$ has order $21$ as soon as $x$ has order $3$ and $y$ has order $7$ (both $x$ and $y$ exist by Cauchy, for example).

For $G$ nonabelian:

  • $Z(G)$ can't have order $3$, $7$, $3.5$, $3.7$, $5.7$, because otherwise $G/Z(G)$ would be cyclic, and hence $G$ abelian: contradiction (for the cases $|Z(G)|=3,7$, note that $5\nmid 7-1$ and $3\nmid 5-1$).
  • If $|Z(G)|=5$, then, for $y$ any element of order $7$ in $G$, the subgroup$^\dagger$ $Z(G)\langle y\rangle$ is normal, because there isn't room enough to accomodate more than one subgroup of order $35$. But then, $\langle y\rangle\unlhd G$ either, and hence $\langle y\rangle\langle x\rangle$ is a subgroup of order $21$ as soon as $x$ is an element of order $3$.
  • if $|Z(G)|=1$, then a subgroup of order $5$ can't be normal, as it can't be obtained as union of conjugacy classes (in fact, $4=5-1$ is not divisable by the size of the smallest nonsingleton conjugacy class, namely $3$). The same holds for a subgroup of order $3$, which is then non-normal either. If there is a normal subgroup of order $7$ (in principle this is allowed, as $7=1+3+3$), say $K$, then again as before there's a subgroup of order $21$ (the set-wise product of $K$ with the group generated by any element of order $3$), and we are done. If there isn't any normal subgroup of order $7$, then I believe we really need Sylow III to rule out such a possibility (all three $n_3$, $n_5$ and $n_7$ greater than $1$). Btw, with Sylow we'd already get $n_5=1$, so actually there is no centerless group of order $105$, and we could consider "virtually proved" the claim with the two previous bullet points for the nonabelian case.

$^\dagger$Recall that $Z(G)\unlhd G$.

citadel
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