Show that each subgroup generated by prime integer is maximal in $(Bbb Z, +)$.
$begingroup$
Show that each subgroup generated by prime integer is maximal in $(Bbb Z, +)$.
Here I know that we can prove maximal by showing its quotient group is simple.
But how can I approach "each subgroup generated by prime integer"?
group-theory prime-numbers
edited Dec 8 '18 at 5:03
Shaun
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
$endgroup$
add a comment |
$begingroup$
Show that each subgroup generated by prime integer is maximal in $(Bbb Z, +)$.
Here I know that we can prove maximal by showing its quotient group is simple.
But how can I approach "each subgroup generated by prime integer"?
group-theory prime-numbers
edited Dec 8 '18 at 5:03
Shaun
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
$endgroup$
$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06
add a comment |
$begingroup$
Show that each subgroup generated by prime integer is maximal in $(Bbb Z, +)$.
Here I know that we can prove maximal by showing its quotient group is simple.
But how can I approach "each subgroup generated by prime integer"?
group-theory prime-numbers
edited Dec 8 '18 at 5:03
Shaun
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
$endgroup$
Show that each subgroup generated by prime integer is maximal in $(Bbb Z, +)$.
Here I know that we can prove maximal by showing its quotient group is simple.
But how can I approach "each subgroup generated by prime integer"?
group-theory prime-numbers
group-theory prime-numbers
edited Dec 8 '18 at 5:03
Shaun
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
edited Dec 8 '18 at 5:03
Shaun
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
edited Dec 8 '18 at 5:03
Shaun
8,929113681
edited Dec 8 '18 at 5:03
Shaun
8,929113681
edited Dec 8 '18 at 5:03
Shaun
8,929113681
8,929113681
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
asked Dec 8 '18 at 4:58
Himanshu Himanshu
134
134
$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06
add a comment |
$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06
$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So if you work out as Shaun mentioned in the comment, you will see that the subgroup generated by the prime number $p $, $$langle p rangle={ldots,-2p,-p,0,p,ldots}=pBbb Z.$$
In fact,since $Bbb Z$ is cyclic, every subgroup of $Bbb Z$ is cyclic,i.e.,every subgroup of $Bbb Z$ is of the form $nBbb Z$ for some $nin Bbb Z$.
To show $pBbb Z$ is maximal in $Bbb Z$, you can use the fact that$$frac{Bbb Z}{pBbb Z}cong Bbb Z_p. $$
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
add a comment |
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$begingroup$
So if you work out as Shaun mentioned in the comment, you will see that the subgroup generated by the prime number $p $, $$langle p rangle={ldots,-2p,-p,0,p,ldots}=pBbb Z.$$
In fact,since $Bbb Z$ is cyclic, every subgroup of $Bbb Z$ is cyclic,i.e.,every subgroup of $Bbb Z$ is of the form $nBbb Z$ for some $nin Bbb Z$.
To show $pBbb Z$ is maximal in $Bbb Z$, you can use the fact that$$frac{Bbb Z}{pBbb Z}cong Bbb Z_p. $$
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
add a comment |
$begingroup$
So if you work out as Shaun mentioned in the comment, you will see that the subgroup generated by the prime number $p $, $$langle p rangle={ldots,-2p,-p,0,p,ldots}=pBbb Z.$$
In fact,since $Bbb Z$ is cyclic, every subgroup of $Bbb Z$ is cyclic,i.e.,every subgroup of $Bbb Z$ is of the form $nBbb Z$ for some $nin Bbb Z$.
To show $pBbb Z$ is maximal in $Bbb Z$, you can use the fact that$$frac{Bbb Z}{pBbb Z}cong Bbb Z_p. $$
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
add a comment |
$begingroup$
So if you work out as Shaun mentioned in the comment, you will see that the subgroup generated by the prime number $p $, $$langle p rangle={ldots,-2p,-p,0,p,ldots}=pBbb Z.$$
In fact,since $Bbb Z$ is cyclic, every subgroup of $Bbb Z$ is cyclic,i.e.,every subgroup of $Bbb Z$ is of the form $nBbb Z$ for some $nin Bbb Z$.
To show $pBbb Z$ is maximal in $Bbb Z$, you can use the fact that$$frac{Bbb Z}{pBbb Z}cong Bbb Z_p. $$
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
So if you work out as Shaun mentioned in the comment, you will see that the subgroup generated by the prime number $p $, $$langle p rangle={ldots,-2p,-p,0,p,ldots}=pBbb Z.$$
In fact,since $Bbb Z$ is cyclic, every subgroup of $Bbb Z$ is cyclic,i.e.,every subgroup of $Bbb Z$ is of the form $nBbb Z$ for some $nin Bbb Z$.
To show $pBbb Z$ is maximal in $Bbb Z$, you can use the fact that$$frac{Bbb Z}{pBbb Z}cong Bbb Z_p. $$
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 5:51
Thomas ShelbyThomas Shelby
2,498221
2,498221
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
add a comment |
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
1
1
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
$begingroup$
I have understood.Thanks
$endgroup$
– Himanshu
Dec 8 '18 at 15:48
add a comment |
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$begingroup$
Check this out math.stackexchange.com/a/2811741
$endgroup$
– marya
Dec 8 '18 at 5:05
$begingroup$
A subgroup generated by an element $g$ of a group $G$ is the subgroup given by all multiples of $g$ and its inverse, all with respect to the operation on $G$.
$endgroup$
– Shaun
Dec 8 '18 at 5:06