Mixing
Why does $a^p equiv 1 (text {mod} q)$?
$begingroup$
Suppose $G$ is a group of order $pq$ with $p<q, p nmid q-1$ and $p,q$ are primes. Then $G$ is cyclic.
The way our instructor has proved this theorem is as follows $:$
He first proved that $G$ has a unique Sylow-$p$-subgroup $S_p$ of order $p$ and a unique Sylow-$q$-subgroup of order $q$ by using Sylow's second theorem. Since conjugation of a Sylow-$p$-subgroup by the elements of $G$ is another Sylow-$p$-subgroup of $G$ so by uniqueness it follows that $S_p$ and $S_q$ are normal subgroups of $G$. Then he takes an element $g$ of order $p$ from $S_p$ and an element $h$ of order $q$ from $S_q$ which are possible to take since both $S_p$ and $S_q$ are cyclic of order $p$ and $q$ respectively. Now using the normality criterion of $S_q$ it follows that $ghg^{-1} in S_q$. Since $S_q$ is a cyclic group generated by $h$ so $ghg^{-1} = h^a$ for some $0 le a le q-1$. At this place he claims that $a^p equiv 1 (text {mod} q)$ as $g^p = e$. Though I don't understand why it should be true. Then the last few arguments go as follows $:$
So $a$ has order dividing $p$ in $(Bbb Z/ q Bbb Z)^*$ which is of order $q-1$. Since $p nmid q-1$ it follows that $a equiv 1 (text {mod} q)$. So $ghg^{-1} = h$ i.e. $g$ and $h$ commute with each other. Since order of $g$ and $h$ are relatively prime to each other it follows that $o(gh) = o(g)o(h) = pq,$ where $o(g)$ denotes the order of the element $g in G$. This proves that $G$ is a cyclic group generated by the element $gh in G$.
Can anybody please help me to point out why $a^p equiv 1 (text {mod} q)$? Then it will be really helpful for me.
Thank you very much.
group-theory proof-explanation cyclic-groups sylow-theory
edited Jan 10 at 14:15
Shaun
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a group of order $pq$ with $p<q, p nmid q-1$ and $p,q$ are primes. Then $G$ is cyclic.
The way our instructor has proved this theorem is as follows $:$
He first proved that $G$ has a unique Sylow-$p$-subgroup $S_p$ of order $p$ and a unique Sylow-$q$-subgroup of order $q$ by using Sylow's second theorem. Since conjugation of a Sylow-$p$-subgroup by the elements of $G$ is another Sylow-$p$-subgroup of $G$ so by uniqueness it follows that $S_p$ and $S_q$ are normal subgroups of $G$. Then he takes an element $g$ of order $p$ from $S_p$ and an element $h$ of order $q$ from $S_q$ which are possible to take since both $S_p$ and $S_q$ are cyclic of order $p$ and $q$ respectively. Now using the normality criterion of $S_q$ it follows that $ghg^{-1} in S_q$. Since $S_q$ is a cyclic group generated by $h$ so $ghg^{-1} = h^a$ for some $0 le a le q-1$. At this place he claims that $a^p equiv 1 (text {mod} q)$ as $g^p = e$. Though I don't understand why it should be true. Then the last few arguments go as follows $:$
So $a$ has order dividing $p$ in $(Bbb Z/ q Bbb Z)^*$ which is of order $q-1$. Since $p nmid q-1$ it follows that $a equiv 1 (text {mod} q)$. So $ghg^{-1} = h$ i.e. $g$ and $h$ commute with each other. Since order of $g$ and $h$ are relatively prime to each other it follows that $o(gh) = o(g)o(h) = pq,$ where $o(g)$ denotes the order of the element $g in G$. This proves that $G$ is a cyclic group generated by the element $gh in G$.
Can anybody please help me to point out why $a^p equiv 1 (text {mod} q)$? Then it will be really helpful for me.
Thank you very much.
group-theory proof-explanation cyclic-groups sylow-theory
edited Jan 10 at 14:15
Shaun
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
$endgroup$
2
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
2
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21
add a comment |
$begingroup$
Suppose $G$ is a group of order $pq$ with $p<q, p nmid q-1$ and $p,q$ are primes. Then $G$ is cyclic.
The way our instructor has proved this theorem is as follows $:$
He first proved that $G$ has a unique Sylow-$p$-subgroup $S_p$ of order $p$ and a unique Sylow-$q$-subgroup of order $q$ by using Sylow's second theorem. Since conjugation of a Sylow-$p$-subgroup by the elements of $G$ is another Sylow-$p$-subgroup of $G$ so by uniqueness it follows that $S_p$ and $S_q$ are normal subgroups of $G$. Then he takes an element $g$ of order $p$ from $S_p$ and an element $h$ of order $q$ from $S_q$ which are possible to take since both $S_p$ and $S_q$ are cyclic of order $p$ and $q$ respectively. Now using the normality criterion of $S_q$ it follows that $ghg^{-1} in S_q$. Since $S_q$ is a cyclic group generated by $h$ so $ghg^{-1} = h^a$ for some $0 le a le q-1$. At this place he claims that $a^p equiv 1 (text {mod} q)$ as $g^p = e$. Though I don't understand why it should be true. Then the last few arguments go as follows $:$
So $a$ has order dividing $p$ in $(Bbb Z/ q Bbb Z)^*$ which is of order $q-1$. Since $p nmid q-1$ it follows that $a equiv 1 (text {mod} q)$. So $ghg^{-1} = h$ i.e. $g$ and $h$ commute with each other. Since order of $g$ and $h$ are relatively prime to each other it follows that $o(gh) = o(g)o(h) = pq,$ where $o(g)$ denotes the order of the element $g in G$. This proves that $G$ is a cyclic group generated by the element $gh in G$.
Can anybody please help me to point out why $a^p equiv 1 (text {mod} q)$? Then it will be really helpful for me.
Thank you very much.
group-theory proof-explanation cyclic-groups sylow-theory
edited Jan 10 at 14:15
Shaun
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
$endgroup$
Suppose $G$ is a group of order $pq$ with $p<q, p nmid q-1$ and $p,q$ are primes. Then $G$ is cyclic.
The way our instructor has proved this theorem is as follows $:$
He first proved that $G$ has a unique Sylow-$p$-subgroup $S_p$ of order $p$ and a unique Sylow-$q$-subgroup of order $q$ by using Sylow's second theorem. Since conjugation of a Sylow-$p$-subgroup by the elements of $G$ is another Sylow-$p$-subgroup of $G$ so by uniqueness it follows that $S_p$ and $S_q$ are normal subgroups of $G$. Then he takes an element $g$ of order $p$ from $S_p$ and an element $h$ of order $q$ from $S_q$ which are possible to take since both $S_p$ and $S_q$ are cyclic of order $p$ and $q$ respectively. Now using the normality criterion of $S_q$ it follows that $ghg^{-1} in S_q$. Since $S_q$ is a cyclic group generated by $h$ so $ghg^{-1} = h^a$ for some $0 le a le q-1$. At this place he claims that $a^p equiv 1 (text {mod} q)$ as $g^p = e$. Though I don't understand why it should be true. Then the last few arguments go as follows $:$
So $a$ has order dividing $p$ in $(Bbb Z/ q Bbb Z)^*$ which is of order $q-1$. Since $p nmid q-1$ it follows that $a equiv 1 (text {mod} q)$. So $ghg^{-1} = h$ i.e. $g$ and $h$ commute with each other. Since order of $g$ and $h$ are relatively prime to each other it follows that $o(gh) = o(g)o(h) = pq,$ where $o(g)$ denotes the order of the element $g in G$. This proves that $G$ is a cyclic group generated by the element $gh in G$.
Can anybody please help me to point out why $a^p equiv 1 (text {mod} q)$? Then it will be really helpful for me.
Thank you very much.
group-theory proof-explanation cyclic-groups sylow-theory
group-theory proof-explanation cyclic-groups sylow-theory
edited Jan 10 at 14:15
Shaun
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
edited Jan 10 at 14:15
Shaun
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
edited Jan 10 at 14:15
Shaun
9,065113682
edited Jan 10 at 14:15
Shaun
9,065113682
edited Jan 10 at 14:15
Shaun
9,065113682
9,065113682
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
asked Jan 9 at 17:19
Dbchatto67Dbchatto67
631116
631116
2
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
2
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21
add a comment |
2
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
2
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21
2
2
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
2
2
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$ghg^{-1} = h^a$, so $g^2hg^{-2} = gh^ag^{-1} = (ghg^{-1})^a = (h^a)^a = h^{a^2}$.
More generally, $g^khg^{-k} = h^{a^k}$. Now $g^p= e$ and $h$ has order $q$ so...
answered Jan 9 at 19:20
MaxMax
14.1k11142
$endgroup$
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
add a comment |
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$begingroup$
$ghg^{-1} = h^a$, so $g^2hg^{-2} = gh^ag^{-1} = (ghg^{-1})^a = (h^a)^a = h^{a^2}$.
More generally, $g^khg^{-k} = h^{a^k}$. Now $g^p= e$ and $h$ has order $q$ so...
answered Jan 9 at 19:20
MaxMax
14.1k11142
$endgroup$
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
add a comment |
$begingroup$
$ghg^{-1} = h^a$, so $g^2hg^{-2} = gh^ag^{-1} = (ghg^{-1})^a = (h^a)^a = h^{a^2}$.
More generally, $g^khg^{-k} = h^{a^k}$. Now $g^p= e$ and $h$ has order $q$ so...
answered Jan 9 at 19:20
MaxMax
14.1k11142
$endgroup$
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
add a comment |
$begingroup$
$ghg^{-1} = h^a$, so $g^2hg^{-2} = gh^ag^{-1} = (ghg^{-1})^a = (h^a)^a = h^{a^2}$.
More generally, $g^khg^{-k} = h^{a^k}$. Now $g^p= e$ and $h$ has order $q$ so...
answered Jan 9 at 19:20
MaxMax
14.1k11142
$endgroup$
$ghg^{-1} = h^a$, so $g^2hg^{-2} = gh^ag^{-1} = (ghg^{-1})^a = (h^a)^a = h^{a^2}$.
More generally, $g^khg^{-k} = h^{a^k}$. Now $g^p= e$ and $h$ has order $q$ so...
answered Jan 9 at 19:20
MaxMax
14.1k11142
answered Jan 9 at 19:20
MaxMax
14.1k11142
answered Jan 9 at 19:20
MaxMax
14.1k11142
answered Jan 9 at 19:20
MaxMax
14.1k11142
14.1k11142
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
add a comment |
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
$begingroup$
Yeah! @Max I have done it myself.
$endgroup$
– Dbchatto67
Jan 10 at 4:52
add a comment |
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2
$begingroup$
There are groups of order $pq$ that are not cyclic.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:20
$begingroup$
$S_3$ has order $6=2times 3$, for instance.
$endgroup$
– lulu
Jan 9 at 17:21
2
$begingroup$
This usually has the condition $q$ not congruent to $1$ mod $p$.
$endgroup$
– Randall
Jan 9 at 17:21