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Given that $$\displaystyle G=\{e,x,x^2,x^3,y,xy,x^2 y,x^3y \}$$ with ${\rm ord}(x)=4$ and ${\rm ord}(y)=2$ such that $xy=yx^3$

Then find the number of elements in the center of the group $G$

The solution I tried:

Given group has order $8=2^3$ thus its the order of its center will be of form $2^n$ and also ${\rm ord}(Z(G))>1$

If the order of $Z(G)$ is $4$ then the quotient ${\rm ord} \dfrac{G}{Z(G)}=2$ which is not possible, and the order $8$ is also not possible because $x$ doesn't commute with every element. So the order of $Z(G)$ will be $2$, but my question is, how can I use the given condition to prove this, what is the use of that condition given in question?

Please help.

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TheStudent
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  • You've shown something stronger, that every non-abelian group of order 8 has 2 elements in it's center. The condition is needed to verify that $G$ is in fact a group of order $8$ (i.e is closed and so every element can be written as $x^i y^j)$ – TAPLON Jan 16 '20 at 16:32
  • so i have to use that condition to show that group is of order $8$? – TheStudent Jan 16 '20 at 16:36
  • I got it, given is that $xy=yx^3$ so multpilying both sides with $x$ we get $xyx=y$ thus it will become $xy=yx^{-1}$ which implies that $x$ and $y$ don not commute ,so order of center is less the $8$ am i right? – TheStudent Jan 16 '20 at 16:39
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    That's correct, but more work than you need. Notice that condition immediately implies $x$ and $y$ do not commute. – TAPLON Jan 16 '20 at 16:49
  • How do you deduce that the center is not trivial? How do you deduce that the quotient can't have size $2$? – Ben Grossmann Jan 16 '20 at 17:29
  • If one is being very pedantic, one could say that it's not given that the order of the group is $8$, as some of the elements could be repeated. (For example, we could have $y=x^2$ and the order conditions would be met, but not the last one. One would then have to check to eliminate such things.) – verret Jan 16 '20 at 18:27
  • @Omnomnomnom that means just looking at group (given above) i can't say that order of group is $8$ ,that means i have to use that given condition to find out the exact order of the group? – TheStudent Jan 17 '20 at 04:36
  • @Student I'm not implying anything like that. The order of a group is how many elements a group has. I'm actually asking: how did you come to the conclusion that the center has more than one element and that it can't have size $2$ before actually figuring out which elements form the center? – Ben Grossmann Jan 17 '20 at 07:39
  • @Student I think you were trying to ask that to the other commenter – Ben Grossmann Jan 17 '20 at 07:40
  • This is a Theorem that if order of group is of form $p^n$ then its center have order >1 ,and for second one i.e size of center is $2$ ,i apologise for that because i think that if $ord (\frac{G}{Z(G)}$ =2 then by Theorem $G$ become cyclic ,but it is not given in question,but i assumed it is given that Group is not cyclic – TheStudent Jan 17 '20 at 07:49

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You correctly identify that $|Z(G)|$ divides $|G|=8$ so $|Z(G)|\in\{1,2,4,8\}$

You correctly note that $|Z(G)|\ne 4$ so you need to show $|Z(G)|\ne 8$ (this follows immediately from the given conditions) and $|Z(G)|\ne 1$. For the latter, just find $1\ne z\in Z(G)$.

Robert Chamberlain
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