Applicable Group Problem Involving Modular Arithmetic
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Question: Show that $(5, 15, 25, 35)$ is a group under multiplication modulo 40.
So I first decided to make a Cayley Table that looks like this:
I apologize that I do not know how to make a Cayley Table in MathJax.
So I know it is closed, and associative because of multiplication inheritance.
I am just confused about the identity element. I could use some descriptive advice about how you find it. I know once you find that, you can use the table to find inverses of each.
abstract-algebra modular-arithmetic
asked Aug 30 at 1:42
Propaloo
404
add a comment |Â
up vote
-1
down vote
favorite
Question: Show that $(5, 15, 25, 35)$ is a group under multiplication modulo 40.
So I first decided to make a Cayley Table that looks like this:
I apologize that I do not know how to make a Cayley Table in MathJax.
So I know it is closed, and associative because of multiplication inheritance.
I am just confused about the identity element. I could use some descriptive advice about how you find it. I know once you find that, you can use the table to find inverses of each.
abstract-algebra modular-arithmetic
asked Aug 30 at 1:42
Propaloo
404
1
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
1
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Question: Show that $(5, 15, 25, 35)$ is a group under multiplication modulo 40.
So I first decided to make a Cayley Table that looks like this:
I apologize that I do not know how to make a Cayley Table in MathJax.
So I know it is closed, and associative because of multiplication inheritance.
I am just confused about the identity element. I could use some descriptive advice about how you find it. I know once you find that, you can use the table to find inverses of each.
abstract-algebra modular-arithmetic
asked Aug 30 at 1:42
Propaloo
404
Question: Show that $(5, 15, 25, 35)$ is a group under multiplication modulo 40.
So I first decided to make a Cayley Table that looks like this:
I apologize that I do not know how to make a Cayley Table in MathJax.
So I know it is closed, and associative because of multiplication inheritance.
I am just confused about the identity element. I could use some descriptive advice about how you find it. I know once you find that, you can use the table to find inverses of each.
abstract-algebra modular-arithmetic
abstract-algebra modular-arithmetic
asked Aug 30 at 1:42
Propaloo
404
asked Aug 30 at 1:42
Propaloo
404
asked Aug 30 at 1:42
Propaloo
404
asked Aug 30 at 1:42
Propaloo
404
asked Aug 30 at 1:42
Propaloo
404
404
1
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
1
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59
add a comment |Â
1
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
1
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59
1
1
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
1
1
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Your identity element is $25$ because in mod $40$ you have
$$ 25times 5=125 equiv 5$$
$$ 25times 15=375equiv 15$$
$$25times 25=625equiv 25$$
$$25times 35=875equiv 35$$
It is interesting to see that for this group every element is its own inverse.
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your identity element is $25$ because in mod $40$ you have
$$ 25times 5=125 equiv 5$$
$$ 25times 15=375equiv 15$$
$$25times 25=625equiv 25$$
$$25times 35=875equiv 35$$
It is interesting to see that for this group every element is its own inverse.
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
Your identity element is $25$ because in mod $40$ you have
$$ 25times 5=125 equiv 5$$
$$ 25times 15=375equiv 15$$
$$25times 25=625equiv 25$$
$$25times 35=875equiv 35$$
It is interesting to see that for this group every element is its own inverse.
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your identity element is $25$ because in mod $40$ you have
$$ 25times 5=125 equiv 5$$
$$ 25times 15=375equiv 15$$
$$25times 25=625equiv 25$$
$$25times 35=875equiv 35$$
It is interesting to see that for this group every element is its own inverse.
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
Your identity element is $25$ because in mod $40$ you have
$$ 25times 5=125 equiv 5$$
$$ 25times 15=375equiv 15$$
$$25times 25=625equiv 25$$
$$25times 35=875equiv 35$$
It is interesting to see that for this group every element is its own inverse.
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 1:57
Mohammad Riazi-Kermani
31k41853
31k41853
add a comment |Â
add a comment |Â
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1
Well, what is the identity element by definition?
– anomaly
Aug 30 at 1:43
I know the identity element of multiplication is just $1$, right?
– Propaloo
Aug 30 at 1:45
1
The identity element of a group is an element $e$ so that $e cdot x = x cdot e = x$ for all $x$ in our group. So is there an element in $(5, 15, 25, 35)$ that when you multiply it by the other elements acts like the identity after you reduce mod $40$?
– JonHales
Aug 30 at 1:47
Ah, interesting. It just clicked! Thank you guys. Sorry, all this stuff is trivial, but new and somewhat challenging to me.
– Propaloo
Aug 30 at 1:59