Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,..)!?

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.