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Prove or disprove: Let $G$ be a group of order 595, then G has a cyclic subgroup of order 35.

What I think is that the statement is true. If $|G|=595=5\times 7\times 17$, then since 5 and 7 divides $|G|$, there exists elements $x,y\in G$ such that $|x|=5, |y|=7$, so $A=\langle x\rangle, B=\langle y\rangle$. By computations we see that number of Sylows 5-subgroups is 1, so there is a unique Sylows 5-subgroup of order 5, which will be $A$, since all Sylows 5-subgroup are conjugate. So, $A$ is normal subgroup of $G$, and $B$ is a subgroup of $G$, Thus, $AB$ is a subgroup of $G$, and

$$|AB|=\frac{|A||B|}{|A\cap B|}=\frac{5\times 7}{1}=35$$ because 5 and 7 are relatively prime.

S0, $|AB|=35$, and $AB\simeq \mathbb Z_{5}\times \mathbb Z_{7} \simeq \mathbb Z_{35}$ since 5 and 7 are relatively prime.

Note: A well known reasult: Any group of order 35 is cyclic, by the Fundamental Theorem of Finite Abelian Groups.

Is my argument correct in general (there are many details, using Sylows Theorem, Cauch Theorem,..)!?

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LaTeX tips: don't use a period for multiplication between numbers: too easy to mistake it with the decimal point or a digit separator; use either \times or \cdot (the latter is a raised dot); instead of < and > for subgroup-generated, use \langle and \rangle; the latter are from the parenthesis family, so they have proper spacing around them, whereas < and > are operators so the space around them is off. – Arturo Magidin Aug 16 '11 at 19:47
You'd better tell us those details. You've left out all the tricky bits. – Chris Eagle Aug 16 '11 at 19:47
It's true that you can find a cyclic group of order $5$ and a cyclic group of order $7$. But you cannot jump from that to saying that there is a subgroup of order $35$: $AB$ need not be a subgroup (for instance, take $G=S_3$: it has a cyclic subgroup of order $2$, a cyclic subgroup of order $3$, but no cyclic subgroup of order $2\times 3=6$). You don't know that $AB\cong A\times B$, because you do not know ahead of time that $x$ and $y$ commute. – Arturo Magidin Aug 16 '11 at 19:49
@Jon: To quote Hendrik Lenstra: "The problem with wrong proofs to correct statements is that it is very hard to give a counterexample." The statement is correct, your proof is just not quite done. – Arturo Magidin Aug 16 '11 at 20:02
@Jon: You cannot invoke a theorem about the structure of abelian groups to prove a group is abelian. If $G$ is a group with $35$ elements, then the number of $7$-Sylow subgroups must be $1$ (since it must divide $35$ and be congruent to $1$ modulo $7$). So $B$ is normal **in $AB$ **. Since $A\cap B = \{1\}$, what can you say about $x^{-1}y^{-1}xy$? Note that $xy=yx$ if and only if $x^{-1}y^{-1}xy = 1$. – Arturo Magidin Aug 16 '11 at 20:07

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