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Is there a $G$ group of order $210$ satisfy following properties:

  1. for some $x,y \in G$, order of $x$ is $5$, order of $y$ is $3$
  2. $x^4y^2=(yx)^6$.

Notes:

  1. Problem is mine and there is no source.

  2. Easly we can show that $(xy)^7 = e$, (here $e$ is unit element of $G$.)

  3. Since least common mutliple of $5,3,7$ is $105$, and $105 \mid 210$; we can search a group $G$ that it's order $210$. But I'm not sure: 'Is there such $G$ group?' So, I need a sharp proof.

scarface
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    The second condition can be rewritten as $x^5y^3 = x(yx)^6y$, or - in view of the first condition - $(xy)^7 = e$, the neutral element of $G$. I would try $G$ as a subgroup of $S_7$, with $x$ a cycle of length 5 and $y$ a cycle of length 3. For example, $x=(12345), y=(567)$; then $xy = (1234567)$ which indeed has order 7, so there's hope. What is the smallest subgroup of $S_7$ containing these two cycles, $x$ and $y$? One obvious observation is that they are both in $A_7$, but that subgroup is still too big. There's some work left, but this is what I would try first. –  May 24 '16 at 16:06
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    According to my calculation with GAP, there is no such group. – Peter May 24 '16 at 16:09

1 Answers1

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Here is a proof that no such group exists. I will use the following results:

1) If $p$ is the least prime divisor of the order of $G$ and a $p$-Sylow $S$ of $G$ is cyclic, then $G$ has a normal $p$-complement $N$. This means $NS=G$ and $N \cap S = \{ 1 \}$.

2) If $G$ has a normal $p$-complement $N$, then $N$ is the subgroup generated by all the elements of order prime to $p$ and $|G|=|N||S|$.

3) A group of order $35$ is abelian.

Now let $G$ be a group of order $210$ with elements $x$ and $y$ satisfying those conditions. The first condition tells you in particular that $x \neq 1$ and $ y \neq 1$. Let $z =yx$. As mathguy said in the comments, $z^7=1$, so the order of $z$ divides $7$. In fact, it has to be $7$, because if $z=1$, then $y=x^{-1}$ which is a contradiction with the fact that $y$ has order $3$ and $x^{-1}$ has order $5$.

The least prime divisor of the order of $G$ is $2$ and since $2$ is the highest power of $2$ that divides $210$, the $2$-Sylows of $G$ are cyclic. Then $G$ has a normal $2$-complement $N$, which has order $105$. Since $N$ is the subgroup of $G$ generated by all the elements of order prime to $2$, it contains $x$, $y$ and $z$.

Now we repeat the argument with $N$. Its least prime divisor is $3$ and $3$ is the highest power of $3$ that divides $105$, so its $3$-Sylows are cyclic. Then $N$ has a normal $3$-complement $K$, which are order $35$. Since $K$ is the subgroup of $N$ generated by all the elements of order prime to $3$, it contains $x$ and $z$.

But now the order of $K$ is $35$, so it is abelian. In particular $xz=zx$, which implies $xy=yx$. Then, using the second condition in the statement, we obtain $x^4y^2 = x^6 y^6$, which implies

$$ 1 = x^2 y^4 = x^2 y \Rightarrow y = x^{-2} $$

And then $ x^{-6} = y^3 = 1$, from where $ x = x^6 x^{-5} = 1$, which is a contradiction. So no such group $G$ exists.

Goa'uld
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