Gallian's Exercise 4.76: a 2008 GRE practice exam question: “exactly two of $x^3$, $x^5$, and $x^9$ are...
$begingroup$
This question appears to be new here according to this and this.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.76 ibid. and it's stated to be a problem from a 2008 GRE practice exam.
I would like to prove this result using the tools given in the textbook prior to this exercise. (A free copy of the book is available online.)
The Question:
If $x$ is an element of a cyclic group of order $15$ and exactly two of $x^3$, $x^5$, and $x^9$ are equal, determine $lvert x^{13}rvert$.
My Attempt:
My first concern is whether it's the order of the group, let's call it $G$, or of the element, called $x$, that is $15$. Nonetheless, here are some thoughts, since $x^{15}=e$ either way:
If $x^3=x^5$, then $x^2=e$, but then either $x=e$ or $lvert xrvert=2$; in the former case, $lvert x^{13}rvert=lvert ervert=0$, but in the latter, $$begin{align}0&=lvert x^{15}rvert\ &=lvert (x^2)^7xrvert \ &=lvert xrvert\ &=2;end{align}$$ thus $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
If $x^3=x^9$, then $x^6=e$, so $e=x^{15}=(x^6)^2x^3=x^3$, which means $x=e$ or $lvert xrvert=3$; thus $lvert x^{13}rvert=0$ as before or $$boxed{begin{align}lvert x^{13}rvert &=lvert (x^3)^4xrvert \ &=lvert xrvert\ &=3.end{align}}$$ (EDIT: But $xneq e$.)
If $x^5=x^9$, then $x^4=e$, but now $e=x^{15}=(x^4)^3x^3=x^3$, so that $x^3=x^4$ implies $x=e$; hence $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
[EDIT 2: The boxed bit is the only answer.]
Is this correct? What would be a better way of determining $lvert x^{13}rvert$?
Please help :)
group-theory proof-verification cyclic-groups
asked Jan 20 at 18:33
ShaunShaun
9,268113684
$endgroup$
add a comment |
$begingroup$
This question appears to be new here according to this and this.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.76 ibid. and it's stated to be a problem from a 2008 GRE practice exam.
I would like to prove this result using the tools given in the textbook prior to this exercise. (A free copy of the book is available online.)
The Question:
If $x$ is an element of a cyclic group of order $15$ and exactly two of $x^3$, $x^5$, and $x^9$ are equal, determine $lvert x^{13}rvert$.
My Attempt:
My first concern is whether it's the order of the group, let's call it $G$, or of the element, called $x$, that is $15$. Nonetheless, here are some thoughts, since $x^{15}=e$ either way:
If $x^3=x^5$, then $x^2=e$, but then either $x=e$ or $lvert xrvert=2$; in the former case, $lvert x^{13}rvert=lvert ervert=0$, but in the latter, $$begin{align}0&=lvert x^{15}rvert\ &=lvert (x^2)^7xrvert \ &=lvert xrvert\ &=2;end{align}$$ thus $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
If $x^3=x^9$, then $x^6=e$, so $e=x^{15}=(x^6)^2x^3=x^3$, which means $x=e$ or $lvert xrvert=3$; thus $lvert x^{13}rvert=0$ as before or $$boxed{begin{align}lvert x^{13}rvert &=lvert (x^3)^4xrvert \ &=lvert xrvert\ &=3.end{align}}$$ (EDIT: But $xneq e$.)
If $x^5=x^9$, then $x^4=e$, but now $e=x^{15}=(x^4)^3x^3=x^3$, so that $x^3=x^4$ implies $x=e$; hence $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
[EDIT 2: The boxed bit is the only answer.]
Is this correct? What would be a better way of determining $lvert x^{13}rvert$?
Please help :)
group-theory proof-verification cyclic-groups
asked Jan 20 at 18:33
ShaunShaun
9,268113684
$endgroup$
1
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
2
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37
add a comment |
$begingroup$
This question appears to be new here according to this and this.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.76 ibid. and it's stated to be a problem from a 2008 GRE practice exam.
I would like to prove this result using the tools given in the textbook prior to this exercise. (A free copy of the book is available online.)
The Question:
If $x$ is an element of a cyclic group of order $15$ and exactly two of $x^3$, $x^5$, and $x^9$ are equal, determine $lvert x^{13}rvert$.
My Attempt:
My first concern is whether it's the order of the group, let's call it $G$, or of the element, called $x$, that is $15$. Nonetheless, here are some thoughts, since $x^{15}=e$ either way:
If $x^3=x^5$, then $x^2=e$, but then either $x=e$ or $lvert xrvert=2$; in the former case, $lvert x^{13}rvert=lvert ervert=0$, but in the latter, $$begin{align}0&=lvert x^{15}rvert\ &=lvert (x^2)^7xrvert \ &=lvert xrvert\ &=2;end{align}$$ thus $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
If $x^3=x^9$, then $x^6=e$, so $e=x^{15}=(x^6)^2x^3=x^3$, which means $x=e$ or $lvert xrvert=3$; thus $lvert x^{13}rvert=0$ as before or $$boxed{begin{align}lvert x^{13}rvert &=lvert (x^3)^4xrvert \ &=lvert xrvert\ &=3.end{align}}$$ (EDIT: But $xneq e$.)
If $x^5=x^9$, then $x^4=e$, but now $e=x^{15}=(x^4)^3x^3=x^3$, so that $x^3=x^4$ implies $x=e$; hence $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
[EDIT 2: The boxed bit is the only answer.]
Is this correct? What would be a better way of determining $lvert x^{13}rvert$?
Please help :)
group-theory proof-verification cyclic-groups
asked Jan 20 at 18:33
ShaunShaun
9,268113684
$endgroup$
This question appears to be new here according to this and this.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.76 ibid. and it's stated to be a problem from a 2008 GRE practice exam.
I would like to prove this result using the tools given in the textbook prior to this exercise. (A free copy of the book is available online.)
The Question:
If $x$ is an element of a cyclic group of order $15$ and exactly two of $x^3$, $x^5$, and $x^9$ are equal, determine $lvert x^{13}rvert$.
My Attempt:
My first concern is whether it's the order of the group, let's call it $G$, or of the element, called $x$, that is $15$. Nonetheless, here are some thoughts, since $x^{15}=e$ either way:
If $x^3=x^5$, then $x^2=e$, but then either $x=e$ or $lvert xrvert=2$; in the former case, $lvert x^{13}rvert=lvert ervert=0$, but in the latter, $$begin{align}0&=lvert x^{15}rvert\ &=lvert (x^2)^7xrvert \ &=lvert xrvert\ &=2;end{align}$$ thus $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
If $x^3=x^9$, then $x^6=e$, so $e=x^{15}=(x^6)^2x^3=x^3$, which means $x=e$ or $lvert xrvert=3$; thus $lvert x^{13}rvert=0$ as before or $$boxed{begin{align}lvert x^{13}rvert &=lvert (x^3)^4xrvert \ &=lvert xrvert\ &=3.end{align}}$$ (EDIT: But $xneq e$.)
If $x^5=x^9$, then $x^4=e$, but now $e=x^{15}=(x^4)^3x^3=x^3$, so that $x^3=x^4$ implies $x=e$; hence $lvert x^{13}rvert=0$. (EDIT: But $xneq e$.)
[EDIT 2: The boxed bit is the only answer.]
Is this correct? What would be a better way of determining $lvert x^{13}rvert$?
Please help :)
group-theory proof-verification cyclic-groups
group-theory proof-verification cyclic-groups
asked Jan 20 at 18:33
ShaunShaun
9,268113684
asked Jan 20 at 18:33
ShaunShaun
9,268113684
edited Jan 20 at 22:11
Shaun
asked Jan 20 at 18:33
ShaunShaun
9,268113684
asked Jan 20 at 18:33
ShaunShaun
9,268113684
asked Jan 20 at 18:33
ShaunShaun
9,268113684
9,268113684
1
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
2
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37
add a comment |
1
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
2
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37
1
1
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
2
2
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $x^3,,x^5,,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=ximplies|x^{13}|=|x|=3$.
edited Jan 20 at 21:13
Shaun
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
$endgroup$
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
add a comment |
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$begingroup$
Since $x^3,,x^5,,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=ximplies|x^{13}|=|x|=3$.
edited Jan 20 at 21:13
Shaun
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
$endgroup$
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
add a comment |
$begingroup$
Since $x^3,,x^5,,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=ximplies|x^{13}|=|x|=3$.
edited Jan 20 at 21:13
Shaun
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
$endgroup$
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
add a comment |
$begingroup$
Since $x^3,,x^5,,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=ximplies|x^{13}|=|x|=3$.
edited Jan 20 at 21:13
Shaun
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
$endgroup$
Since $x^3,,x^5,,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=ximplies|x^{13}|=|x|=3$.
edited Jan 20 at 21:13
Shaun
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
edited Jan 20 at 21:13
Shaun
9,268113684
edited Jan 20 at 21:13
Shaun
9,268113684
edited Jan 20 at 21:13
Shaun
9,268113684
9,268113684
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
answered Jan 20 at 19:12
J.G.J.G.
27.5k22843
27.5k22843
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
add a comment |
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
$begingroup$
This seems very slick. Thank you. How did you get $lvert xrvert=3$ though?
$endgroup$
– Shaun
Jan 20 at 21:26
1
1
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
$begingroup$
@Shaun $|x|le 2$ contradicts the setup, and $3$ if the only other common factor of $12$ and $15$.
$endgroup$
– J.G.
Jan 20 at 21:29
1
1
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
$begingroup$
Thank you. That's very clear now :)
$endgroup$
– Shaun
Jan 20 at 21:31
add a comment |
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1
$begingroup$
We should interpret that the order of the group is $15$. The properties asserted for $x$ rule out that it could have order $15$. Of course $x$ will generate a subgroup of the same order as $x$.
$endgroup$
– hardmath
Jan 20 at 18:36
2
$begingroup$
I've just noticed that the "exactly two" specification of the powers eliminates the $x=e$ cases.
$endgroup$
– Shaun
Jan 20 at 18:37