Mixing
Calculating the order of an power element in a group
Trying to understand the group order material. While practising, I came across with the following question.
Consider $gin G$ so $o(g)=15$. calculate $o(g^7),o(g^{-1})$.
I'm not sure how to approach this question. I understand from $o(g)=15$ that $g^{15}=e$. So in order to calculate $o(g^5)$ we will have to do some arithmetic on $g^{15}$. we need to find $kinmathbb{N}$ so $(g^7)^k = e$. We are getting $g^{7k}=e$ and now I'm not sure what do next. Also probably the bigger problem is to understand how to calculate $o(g^{-1})$.
group-theory
asked Nov 20 '18 at 19:03
vesii
906
add a comment |
Trying to understand the group order material. While practising, I came across with the following question.
Consider $gin G$ so $o(g)=15$. calculate $o(g^7),o(g^{-1})$.
I'm not sure how to approach this question. I understand from $o(g)=15$ that $g^{15}=e$. So in order to calculate $o(g^5)$ we will have to do some arithmetic on $g^{15}$. we need to find $kinmathbb{N}$ so $(g^7)^k = e$. We are getting $g^{7k}=e$ and now I'm not sure what do next. Also probably the bigger problem is to understand how to calculate $o(g^{-1})$.
group-theory
asked Nov 20 '18 at 19:03
vesii
906
When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12
add a comment |
Trying to understand the group order material. While practising, I came across with the following question.
Consider $gin G$ so $o(g)=15$. calculate $o(g^7),o(g^{-1})$.
I'm not sure how to approach this question. I understand from $o(g)=15$ that $g^{15}=e$. So in order to calculate $o(g^5)$ we will have to do some arithmetic on $g^{15}$. we need to find $kinmathbb{N}$ so $(g^7)^k = e$. We are getting $g^{7k}=e$ and now I'm not sure what do next. Also probably the bigger problem is to understand how to calculate $o(g^{-1})$.
group-theory
asked Nov 20 '18 at 19:03
vesii
906
Trying to understand the group order material. While practising, I came across with the following question.
Consider $gin G$ so $o(g)=15$. calculate $o(g^7),o(g^{-1})$.
I'm not sure how to approach this question. I understand from $o(g)=15$ that $g^{15}=e$. So in order to calculate $o(g^5)$ we will have to do some arithmetic on $g^{15}$. we need to find $kinmathbb{N}$ so $(g^7)^k = e$. We are getting $g^{7k}=e$ and now I'm not sure what do next. Also probably the bigger problem is to understand how to calculate $o(g^{-1})$.
group-theory
group-theory
asked Nov 20 '18 at 19:03
vesii
906
asked Nov 20 '18 at 19:03
vesii
906
edited Nov 20 '18 at 19:30
asked Nov 20 '18 at 19:03
vesii
906
asked Nov 20 '18 at 19:03
vesii
906
asked Nov 20 '18 at 19:03
vesii
906
906
When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12
add a comment |
When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12
When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12
add a comment |
2 Answers
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The assertion $operatorname{ord}(g)=12$ means two things:
$g^{12}=e$;- if $kin{1,2,ldots,11}$, then $g^kneq e$.
A standard fact about the order of an element is this: $g^k=eimpliesoperatorname{ord}(g)mid k$.
From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5times12}=(g^{12})^5=e^5=e.$$Therefore, $operatorname{ord}(g^5)$ is at most $12$.
Now, take $kin{1,2,ldots,11}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $operatorname{ord}(g)=12$, $12mid5k$. Since $gcd(12,5)=1$, it follows that $12mid k$, which is impossible, since $kin{1,2,ldots,11}$. This proves that $operatorname{ord}(g^5)=12$.
Now, prove that $operatorname{ord}(g^{-1})$ is also equal to $12$.
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
add a comment |
If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:
An element of a group has the same order as its inverse
For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
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active
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votes
active
oldest
votes
The assertion $operatorname{ord}(g)=12$ means two things:
$g^{12}=e$;- if $kin{1,2,ldots,11}$, then $g^kneq e$.
A standard fact about the order of an element is this: $g^k=eimpliesoperatorname{ord}(g)mid k$.
From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5times12}=(g^{12})^5=e^5=e.$$Therefore, $operatorname{ord}(g^5)$ is at most $12$.
Now, take $kin{1,2,ldots,11}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $operatorname{ord}(g)=12$, $12mid5k$. Since $gcd(12,5)=1$, it follows that $12mid k$, which is impossible, since $kin{1,2,ldots,11}$. This proves that $operatorname{ord}(g^5)=12$.
Now, prove that $operatorname{ord}(g^{-1})$ is also equal to $12$.
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
add a comment |
The assertion $operatorname{ord}(g)=12$ means two things:
$g^{12}=e$;- if $kin{1,2,ldots,11}$, then $g^kneq e$.
A standard fact about the order of an element is this: $g^k=eimpliesoperatorname{ord}(g)mid k$.
From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5times12}=(g^{12})^5=e^5=e.$$Therefore, $operatorname{ord}(g^5)$ is at most $12$.
Now, take $kin{1,2,ldots,11}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $operatorname{ord}(g)=12$, $12mid5k$. Since $gcd(12,5)=1$, it follows that $12mid k$, which is impossible, since $kin{1,2,ldots,11}$. This proves that $operatorname{ord}(g^5)=12$.
Now, prove that $operatorname{ord}(g^{-1})$ is also equal to $12$.
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
add a comment |
The assertion $operatorname{ord}(g)=12$ means two things:
$g^{12}=e$;- if $kin{1,2,ldots,11}$, then $g^kneq e$.
A standard fact about the order of an element is this: $g^k=eimpliesoperatorname{ord}(g)mid k$.
From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5times12}=(g^{12})^5=e^5=e.$$Therefore, $operatorname{ord}(g^5)$ is at most $12$.
Now, take $kin{1,2,ldots,11}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $operatorname{ord}(g)=12$, $12mid5k$. Since $gcd(12,5)=1$, it follows that $12mid k$, which is impossible, since $kin{1,2,ldots,11}$. This proves that $operatorname{ord}(g^5)=12$.
Now, prove that $operatorname{ord}(g^{-1})$ is also equal to $12$.
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
The assertion $operatorname{ord}(g)=12$ means two things:
$g^{12}=e$;- if $kin{1,2,ldots,11}$, then $g^kneq e$.
A standard fact about the order of an element is this: $g^k=eimpliesoperatorname{ord}(g)mid k$.
From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5times12}=(g^{12})^5=e^5=e.$$Therefore, $operatorname{ord}(g^5)$ is at most $12$.
Now, take $kin{1,2,ldots,11}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $operatorname{ord}(g)=12$, $12mid5k$. Since $gcd(12,5)=1$, it follows that $12mid k$, which is impossible, since $kin{1,2,ldots,11}$. This proves that $operatorname{ord}(g^5)=12$.
Now, prove that $operatorname{ord}(g^{-1})$ is also equal to $12$.
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
answered Nov 20 '18 at 19:14
José Carlos Santos
151k22123224
151k22123224
add a comment |
add a comment |
If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:
An element of a group has the same order as its inverse
For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
add a comment |
If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:
An element of a group has the same order as its inverse
For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
add a comment |
If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:
An element of a group has the same order as its inverse
For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:
An element of a group has the same order as its inverse
For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
answered Nov 20 '18 at 19:11
Dietrich Burde
77.8k64386
77.8k64386
add a comment |
add a comment |
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When does $12mid5k$? For $g^{-1}$ it's even simpler... $g^{-k}=e$ iff $g^k=e$
– Federico
Nov 20 '18 at 19:09
@Federico Thanks for the replay. why $12|5k$?
– vesii
Nov 20 '18 at 19:11
$g^k=e$ iff $12mid k$. Now, to find when $g^{5k}=e$, you have to find when $12mid5k$.
– Federico
Nov 20 '18 at 19:12