Homomorphism: How do we get the equality?
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Let $Z = (mathbbZ,+)$ the additive group of integers and $G = (M,star )$ an arbitrary group.
I want to show that for all $a in G$ the map $phi_a : Z rightarrow G$ defined by $phi_a(k) = a^k$ is an homomorphism from $Z$ to $G$.
Let $m,nin Z$. Then we have that $phi_a(m+n)=a^m+n$ and $phi_a(m)star phi_a(n)=a^mstar a^n$.
Do we consider $star$ as a multiplication and so $a^m+n=a^mstar a^n$, or how do we get the equality?
abstract-algebra group-homomorphism
asked Aug 26 at 22:56
Mary Star
2,82982056
add a comment |Â
up vote
1
down vote
favorite
Let $Z = (mathbbZ,+)$ the additive group of integers and $G = (M,star )$ an arbitrary group.
I want to show that for all $a in G$ the map $phi_a : Z rightarrow G$ defined by $phi_a(k) = a^k$ is an homomorphism from $Z$ to $G$.
Let $m,nin Z$. Then we have that $phi_a(m+n)=a^m+n$ and $phi_a(m)star phi_a(n)=a^mstar a^n$.
Do we consider $star$ as a multiplication and so $a^m+n=a^mstar a^n$, or how do we get the equality?
abstract-algebra group-homomorphism
asked Aug 26 at 22:56
Mary Star
2,82982056
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $Z = (mathbbZ,+)$ the additive group of integers and $G = (M,star )$ an arbitrary group.
I want to show that for all $a in G$ the map $phi_a : Z rightarrow G$ defined by $phi_a(k) = a^k$ is an homomorphism from $Z$ to $G$.
Let $m,nin Z$. Then we have that $phi_a(m+n)=a^m+n$ and $phi_a(m)star phi_a(n)=a^mstar a^n$.
Do we consider $star$ as a multiplication and so $a^m+n=a^mstar a^n$, or how do we get the equality?
abstract-algebra group-homomorphism
asked Aug 26 at 22:56
Mary Star
2,82982056
Let $Z = (mathbbZ,+)$ the additive group of integers and $G = (M,star )$ an arbitrary group.
I want to show that for all $a in G$ the map $phi_a : Z rightarrow G$ defined by $phi_a(k) = a^k$ is an homomorphism from $Z$ to $G$.
Let $m,nin Z$. Then we have that $phi_a(m+n)=a^m+n$ and $phi_a(m)star phi_a(n)=a^mstar a^n$.
Do we consider $star$ as a multiplication and so $a^m+n=a^mstar a^n$, or how do we get the equality?
abstract-algebra group-homomorphism
asked Aug 26 at 22:56
Mary Star
2,82982056
asked Aug 26 at 22:56
Mary Star
2,82982056
asked Aug 26 at 22:56
Mary Star
2,82982056
asked Aug 26 at 22:56
Mary Star
2,82982056
2,82982056
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The question as phrased is leaving it to you to figure out that the notation $a^n$ means $overbracea star ldots star a^mbox$n$ $a$s$ . I.e. $a^n$ is defined by:
$$
beginalign
a^0 &= e \
a^n+1 &= a^n star a
endalign
$$
where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m star a^n = a^m+n$.
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
1
down vote
Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m star a^n = a^m+n$ is given as the associative property of the group operation.
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The question as phrased is leaving it to you to figure out that the notation $a^n$ means $overbracea star ldots star a^mbox$n$ $a$s$ . I.e. $a^n$ is defined by:
$$
beginalign
a^0 &= e \
a^n+1 &= a^n star a
endalign
$$
where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m star a^n = a^m+n$.
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
2
down vote
accepted
The question as phrased is leaving it to you to figure out that the notation $a^n$ means $overbracea star ldots star a^mbox$n$ $a$s$ . I.e. $a^n$ is defined by:
$$
beginalign
a^0 &= e \
a^n+1 &= a^n star a
endalign
$$
where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m star a^n = a^m+n$.
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The question as phrased is leaving it to you to figure out that the notation $a^n$ means $overbracea star ldots star a^mbox$n$ $a$s$ . I.e. $a^n$ is defined by:
$$
beginalign
a^0 &= e \
a^n+1 &= a^n star a
endalign
$$
where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m star a^n = a^m+n$.
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
The question as phrased is leaving it to you to figure out that the notation $a^n$ means $overbracea star ldots star a^mbox$n$ $a$s$ . I.e. $a^n$ is defined by:
$$
beginalign
a^0 &= e \
a^n+1 &= a^n star a
endalign
$$
where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m star a^n = a^m+n$.
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
answered Aug 26 at 23:29
Rob Arthan
27.4k42865
27.4k42865
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
Thank you!! :-)
– Mary Star
2 days ago
Thank you!! :-)
– Mary Star
2 days ago
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
1
down vote
Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m star a^n = a^m+n$ is given as the associative property of the group operation.
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
1
down vote
Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m star a^n = a^m+n$ is given as the associative property of the group operation.
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m star a^n = a^m+n$ is given as the associative property of the group operation.
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m star a^n = a^m+n$ is given as the associative property of the group operation.
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
answered Aug 26 at 23:08
Mohammad Riazi-Kermani
30.6k41852
30.6k41852
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
Thank you!! :-)
– Mary Star
2 days ago
Thank you!! :-)
– Mary Star
2 days ago
Thank you!! :-)
– Mary Star
2 days ago
add a comment |Â
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