Mixing
Confused at to why $prod_{g in G}g^2=e$
$begingroup$
Let $(G, circ)$ be a finite abelian group with neutral element $e$.
$1.$ Show that in general $prod_{g in G}g^nneq e$ for $n=1$
My idea:
I used $mathbb Z/2mathbb Z$, it follows that $prod_{g in G}g=[0]+[1]neq[0] in mathbb Z/2mathbb Z$
$2.$ Show that there exists an $n in mathbb N$ so that $prod_{g in G}g^n=e$
I found a solution but I do not understand it. It states
$prod_{g in G}g^2=prod_{g in G}(gcdot g^{-1})$
I do not understand this step... Why is $g^{2}=gcdot g^{-1}$?
Thank for your help
group-theory finite-groups abelian-groups
edited Jan 31 at 16:00
Shaun
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
$endgroup$
add a comment |
$begingroup$
Let $(G, circ)$ be a finite abelian group with neutral element $e$.
$1.$ Show that in general $prod_{g in G}g^nneq e$ for $n=1$
My idea:
I used $mathbb Z/2mathbb Z$, it follows that $prod_{g in G}g=[0]+[1]neq[0] in mathbb Z/2mathbb Z$
$2.$ Show that there exists an $n in mathbb N$ so that $prod_{g in G}g^n=e$
I found a solution but I do not understand it. It states
$prod_{g in G}g^2=prod_{g in G}(gcdot g^{-1})$
I do not understand this step... Why is $g^{2}=gcdot g^{-1}$?
Thank for your help
group-theory finite-groups abelian-groups
edited Jan 31 at 16:00
Shaun
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
$endgroup$
$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
1
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25
add a comment |
$begingroup$
Let $(G, circ)$ be a finite abelian group with neutral element $e$.
$1.$ Show that in general $prod_{g in G}g^nneq e$ for $n=1$
My idea:
I used $mathbb Z/2mathbb Z$, it follows that $prod_{g in G}g=[0]+[1]neq[0] in mathbb Z/2mathbb Z$
$2.$ Show that there exists an $n in mathbb N$ so that $prod_{g in G}g^n=e$
I found a solution but I do not understand it. It states
$prod_{g in G}g^2=prod_{g in G}(gcdot g^{-1})$
I do not understand this step... Why is $g^{2}=gcdot g^{-1}$?
Thank for your help
group-theory finite-groups abelian-groups
edited Jan 31 at 16:00
Shaun
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
$endgroup$
Let $(G, circ)$ be a finite abelian group with neutral element $e$.
$1.$ Show that in general $prod_{g in G}g^nneq e$ for $n=1$
My idea:
I used $mathbb Z/2mathbb Z$, it follows that $prod_{g in G}g=[0]+[1]neq[0] in mathbb Z/2mathbb Z$
$2.$ Show that there exists an $n in mathbb N$ so that $prod_{g in G}g^n=e$
I found a solution but I do not understand it. It states
$prod_{g in G}g^2=prod_{g in G}(gcdot g^{-1})$
I do not understand this step... Why is $g^{2}=gcdot g^{-1}$?
Thank for your help
group-theory finite-groups abelian-groups
group-theory finite-groups abelian-groups
edited Jan 31 at 16:00
Shaun
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
edited Jan 31 at 16:00
Shaun
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
edited Jan 31 at 16:00
Shaun
10.3k113686
edited Jan 31 at 16:00
Shaun
10.3k113686
edited Jan 31 at 16:00
Shaun
10.3k113686
10.3k113686
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
asked Jan 31 at 14:31
MinaThumaMinaThuma
2709
2709
$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
1
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25
add a comment |
$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
1
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25
$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
1
1
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not true that $g^2 = g circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write
begin{align}
prod_{g in G} g^2 &= prod_{g in G}g circ (g^{-1})^{-1}\
&= prod_{g in G} g times prod_{g in G} (g^{-1})^{-1} \
&= prod_{g in G} g times prod_{h in G} h^{-1} \
&= prod_{g in G} g times prod_{g in G} g^{-1} \
&= prod_{g in G}g circ g^{-1}.
end{align}
edited Jan 31 at 15:57
Shaun
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
add a comment |
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1 Answer
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$begingroup$
It is not true that $g^2 = g circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write
begin{align}
prod_{g in G} g^2 &= prod_{g in G}g circ (g^{-1})^{-1}\
&= prod_{g in G} g times prod_{g in G} (g^{-1})^{-1} \
&= prod_{g in G} g times prod_{h in G} h^{-1} \
&= prod_{g in G} g times prod_{g in G} g^{-1} \
&= prod_{g in G}g circ g^{-1}.
end{align}
edited Jan 31 at 15:57
Shaun
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
add a comment |
$begingroup$
It is not true that $g^2 = g circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write
begin{align}
prod_{g in G} g^2 &= prod_{g in G}g circ (g^{-1})^{-1}\
&= prod_{g in G} g times prod_{g in G} (g^{-1})^{-1} \
&= prod_{g in G} g times prod_{h in G} h^{-1} \
&= prod_{g in G} g times prod_{g in G} g^{-1} \
&= prod_{g in G}g circ g^{-1}.
end{align}
edited Jan 31 at 15:57
Shaun
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
add a comment |
$begingroup$
It is not true that $g^2 = g circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write
begin{align}
prod_{g in G} g^2 &= prod_{g in G}g circ (g^{-1})^{-1}\
&= prod_{g in G} g times prod_{g in G} (g^{-1})^{-1} \
&= prod_{g in G} g times prod_{h in G} h^{-1} \
&= prod_{g in G} g times prod_{g in G} g^{-1} \
&= prod_{g in G}g circ g^{-1}.
end{align}
edited Jan 31 at 15:57
Shaun
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
$endgroup$
It is not true that $g^2 = g circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write
begin{align}
prod_{g in G} g^2 &= prod_{g in G}g circ (g^{-1})^{-1}\
&= prod_{g in G} g times prod_{g in G} (g^{-1})^{-1} \
&= prod_{g in G} g times prod_{h in G} h^{-1} \
&= prod_{g in G} g times prod_{g in G} g^{-1} \
&= prod_{g in G}g circ g^{-1}.
end{align}
edited Jan 31 at 15:57
Shaun
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
edited Jan 31 at 15:57
Shaun
10.3k113686
edited Jan 31 at 15:57
Shaun
10.3k113686
edited Jan 31 at 15:57
Shaun
10.3k113686
10.3k113686
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
answered Jan 31 at 14:37
Mees de VriesMees de Vries
17.6k13060
17.6k13060
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
add a comment |
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
$begingroup$
@MinaThuma, in the second final step I am reordering the product to put the two products back together (and arguably I am changing the ordering in step three or four, although that is hidden in the notation; really $prod_{g in G} g$ does not even make sense for non-abelian groups without fixing an ordering on $G$).
$endgroup$
– Mees de Vries
Jan 31 at 15:26
add a comment |
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$begingroup$
The group is abelian so you can rearrange the order of the products. Than in a group an element has a unique inverse (which can be also itself). Try using these two properties to show the equality that you need...
$endgroup$
– Thomas
Jan 31 at 14:35
$begingroup$
Note that the solution does not claim that for each $gin G$, $g^2 = gcdot g^{-1}$. It only claims that the two products are equal.
$endgroup$
– rogerl
Jan 31 at 14:36
$begingroup$
It is not true in general that $g^2=g cdot g^{-1}$, but if you take the product of all elements twice you will have each element and its inverse.
$endgroup$
– J. W. Tanner
Jan 31 at 14:36
$begingroup$
Note that the second question can be answered easily by taking $n=|G|$ and applying Lagrange's theorem.
$endgroup$
– lisyarus
Jan 31 at 14:47
1
$begingroup$
Or even more easily by taking $n = prod_{g in G} mathrm{ord}(g)$.
$endgroup$
– Mees de Vries
Jan 31 at 15:25