2
$\begingroup$

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,..)!?

  • 0
    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.2011-08-16
  • 0
    You'd better tell us those details. You've left out all the tricky bits.2011-08-16
  • 0
    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.2011-08-16
  • 0
    Remember AB isn't always a subgroup. It is, however if you can show that one of A or B has to be normal. Hint: How many Sylow 5-subgroups does there have to be?2011-08-16
  • 0
    I provided the missing details above.2011-08-16
  • 0
    @Jon: You aren't done, though. You can have two cyclic subgroups $A$ and $B$, with $AB$ a subgroup, yet $AB$ not isomorphic to $A\times B$: again, consider $G=S_3$ with $A$ the normal subgroup of order $3$ and $B$ a subgroup of order $2$. You would still need to show that $xy=yx$ to conclude that $AB\cong A\times B$.2011-08-16
  • 0
    OK, so you mean that the satement above is not correct!2011-08-16
  • 0
    You should show your computations showing there is only one Sylow 5-subgroup, and explain why a group of order 35 has to be cyclic. Another hint: in Arturo's example, 2 | (3 - 1)2011-08-16
  • 2
    @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.2011-08-16
  • 0
    You can't invoke the Fundamental Theorem of Ab Grps, since you don't know the group is abelian.2011-08-16
  • 0
    @ Matt, " explain why a group of order 35 has to be cyclic." : this is another story, but I'm 100% sure it is correct, don't worry about that.2011-08-16
  • 0
    @Jon I'm not worried, but it is a little more subtle than it seems. Is a group of order 15 necessarily cyclic? What about a group of order 21?2011-08-16
  • 3
    @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 $x^{-1}y^{-1}xy = 1$.2011-08-16
  • 0
    Thats right, thank you. I just need to know that the statement is correct or not, I can check the details myself.2011-08-16
  • 0
    @Jon: The original statement is correct, as is the claim that a group of order 35 must be cyclic. And most of what you write is also correct. The only glaring error is your justification for "Every group of order 35 is cyclic." It does not follow from the Fundamental Theorem of Finite Abelian Groups, but rather from other considerations.2011-08-16

0 Answers 0