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How order of an element and it's inverse are equal [closed]
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In group Theory how to prove that an element and it's inverse has same order
group-theory
asked Jul 18 at 17:22
Koushik
11
closed as off-topic by amWhy, Derek Holt, José Carlos Santos, rschwieb, Crostul Jul 18 at 17:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Derek Holt, Jos̩ Carlos Santos, Crostul
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up vote
-2
down vote
favorite
In group Theory how to prove that an element and it's inverse has same order
group-theory
asked Jul 18 at 17:22
Koushik
11
closed as off-topic by amWhy, Derek Holt, José Carlos Santos, rschwieb, Crostul Jul 18 at 17:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Derek Holt, Jos̩ Carlos Santos, Crostul
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
In group Theory how to prove that an element and it's inverse has same order
group-theory
asked Jul 18 at 17:22
Koushik
11
In group Theory how to prove that an element and it's inverse has same order
group-theory
asked Jul 18 at 17:22
Koushik
11
asked Jul 18 at 17:22
Koushik
11
asked Jul 18 at 17:22
Koushik
11
asked Jul 18 at 17:22
Koushik
11
11
closed as off-topic by amWhy, Derek Holt, José Carlos Santos, rschwieb, Crostul Jul 18 at 17:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Derek Holt, Jos̩ Carlos Santos, Crostul
closed as off-topic by amWhy, Derek Holt, José Carlos Santos, rschwieb, Crostul Jul 18 at 17:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Derek Holt, Jos̩ Carlos Santos, Crostul
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38
add a comment |Â
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38
add a comment |Â
1 Answer
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Let $n$ be the order of $g$. Then $g^n=e$, which we can imagine being written as
beginalign*
overbracegldots g^n = e &implies overbracegldots g^n-1 = g^-1 \
&vdots \
&implies g = overbraceg^-1ldots g^-1^n-1\
&implies e = overbraceg^-1ldots g^-1^n
endalign*
so $(g^-1)^n = e$. To complete the proof you have to show that $n$ is the smallest number $k$ such that $(g^-1)^k=e$. Can you finish it from here?
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $n$ be the order of $g$. Then $g^n=e$, which we can imagine being written as
beginalign*
overbracegldots g^n = e &implies overbracegldots g^n-1 = g^-1 \
&vdots \
&implies g = overbraceg^-1ldots g^-1^n-1\
&implies e = overbraceg^-1ldots g^-1^n
endalign*
so $(g^-1)^n = e$. To complete the proof you have to show that $n$ is the smallest number $k$ such that $(g^-1)^k=e$. Can you finish it from here?
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
add a comment |Â
up vote
1
down vote
Let $n$ be the order of $g$. Then $g^n=e$, which we can imagine being written as
beginalign*
overbracegldots g^n = e &implies overbracegldots g^n-1 = g^-1 \
&vdots \
&implies g = overbraceg^-1ldots g^-1^n-1\
&implies e = overbraceg^-1ldots g^-1^n
endalign*
so $(g^-1)^n = e$. To complete the proof you have to show that $n$ is the smallest number $k$ such that $(g^-1)^k=e$. Can you finish it from here?
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $n$ be the order of $g$. Then $g^n=e$, which we can imagine being written as
beginalign*
overbracegldots g^n = e &implies overbracegldots g^n-1 = g^-1 \
&vdots \
&implies g = overbraceg^-1ldots g^-1^n-1\
&implies e = overbraceg^-1ldots g^-1^n
endalign*
so $(g^-1)^n = e$. To complete the proof you have to show that $n$ is the smallest number $k$ such that $(g^-1)^k=e$. Can you finish it from here?
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
Let $n$ be the order of $g$. Then $g^n=e$, which we can imagine being written as
beginalign*
overbracegldots g^n = e &implies overbracegldots g^n-1 = g^-1 \
&vdots \
&implies g = overbraceg^-1ldots g^-1^n-1\
&implies e = overbraceg^-1ldots g^-1^n
endalign*
so $(g^-1)^n = e$. To complete the proof you have to show that $n$ is the smallest number $k$ such that $(g^-1)^k=e$. Can you finish it from here?
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
answered Jul 18 at 17:27
Daniel Littlewood
12.7k11742
12.7k11742
add a comment |Â
add a comment |Â
Use the fact that $(a^-1)^n = (a^n)^-1$.
– Alex Provost
Jul 18 at 17:28
Possible duplicate of An element of a group has the same order as its inverse
– rschwieb
Jul 18 at 17:38
In the future, type the keywords of your question into the search first. Using your title, I think you would immediately see the duplicate I suggested.
– rschwieb
Jul 18 at 17:38