Proving a direct product of groups is a group












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I am trying to prove the following and am looking toward the math.stackexchange community to comment on whether I am on the right track or not. Thank you in advance.




Let $n geq 1$ be an integer. Suppose ${G_1,...,G_n}$ is a finite set of groups. Define the set $G = G_1 times G_2 times ... times G_n$ of ordered n-tuples $(g_1,g_2,...,g_n)$ where $g_i in G_i$. Show that the operation $bullet$: $G times G rightarrow G$ given by $(g_1,g_2,...,g_n) bullet (g'_1,g'_2,...,g'_n) = (g_1g'_1,g_2g'_2,...,g_ng'_n)$ makes $G$ into a group.




Proof of associativity: $((g_1,g_2,...,g_n) bullet (g'_1,g'_2,...g'_n)) bullet (g''_1,g''_2,...,g''_n)$ = $(g_1g'_1,g_2g'_2,...,g_ng'_n) bullet (g''_1,g''_2,...,g''_n) $ = $(g_1g'_1g''_1,g_2g'_2g''_2,...,g_ng'_ng''_n)$ = $(g_1,g_2,...,g_n) bullet (g'_1g''_1,g'_2g''_2,...,g'_ng''_n) = (g_1,g_2,...,g_n) bullet ((g'_1,g'_2,...,g'_n) bullet (g''_1,g''_2,...,g''_n))$





Proof of the existence of an identity For each $G_i in G$, take its identity element $e_i$ and multiply it as follows: $(g_1,g_2,...,g_n) bullet (e_1,e_2,...,e_3)$ = $(g_1,g_2,...,g_n)$.





Proof of the existence of an inverse: Since each $G_i in G$ is a group, simply takes its multiplicative inverse $g_i^{-1}$ and multiply it: $(g_1,g_2,...,g_n) bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.






proof-verification finite-groups direct-product







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  • 1




    $begingroup$
    Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
    $endgroup$
    – Timbuc
    Feb 24 '15 at 6:27












  • $begingroup$
    Thank you very much Timbuc for that tip on associativity.
    $endgroup$
    – letsmakemuffinstogether
    Feb 25 '15 at 3:28
















1












$begingroup$


I am trying to prove the following and am looking toward the math.stackexchange community to comment on whether I am on the right track or not. Thank you in advance.




Let $n geq 1$ be an integer. Suppose ${G_1,...,G_n}$ is a finite set of groups. Define the set $G = G_1 times G_2 times ... times G_n$ of ordered n-tuples $(g_1,g_2,...,g_n)$ where $g_i in G_i$. Show that the operation $bullet$: $G times G rightarrow G$ given by $(g_1,g_2,...,g_n) bullet (g'_1,g'_2,...,g'_n) = (g_1g'_1,g_2g'_2,...,g_ng'_n)$ makes $G$ into a group.




Proof of associativity: $((g_1,g_2,...,g_n) bullet (g'_1,g'_2,...g'_n)) bullet (g''_1,g''_2,...,g''_n)$ = $(g_1g'_1,g_2g'_2,...,g_ng'_n) bullet (g''_1,g''_2,...,g''_n) $ = $(g_1g'_1g''_1,g_2g'_2g''_2,...,g_ng'_ng''_n)$ = $(g_1,g_2,...,g_n) bullet (g'_1g''_1,g'_2g''_2,...,g'_ng''_n) = (g_1,g_2,...,g_n) bullet ((g'_1,g'_2,...,g'_n) bullet (g''_1,g''_2,...,g''_n))$





Proof of the existence of an identity For each $G_i in G$, take its identity element $e_i$ and multiply it as follows: $(g_1,g_2,...,g_n) bullet (e_1,e_2,...,e_3)$ = $(g_1,g_2,...,g_n)$.





Proof of the existence of an inverse: Since each $G_i in G$ is a group, simply takes its multiplicative inverse $g_i^{-1}$ and multiply it: $(g_1,g_2,...,g_n) bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.






proof-verification finite-groups direct-product







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
    $endgroup$
    – Timbuc
    Feb 24 '15 at 6:27












  • $begingroup$
    Thank you very much Timbuc for that tip on associativity.
    $endgroup$
    – letsmakemuffinstogether
    Feb 25 '15 at 3:28














1












1








1





$begingroup$


I am trying to prove the following and am looking toward the math.stackexchange community to comment on whether I am on the right track or not. Thank you in advance.




Let $n geq 1$ be an integer. Suppose ${G_1,...,G_n}$ is a finite set of groups. Define the set $G = G_1 times G_2 times ... times G_n$ of ordered n-tuples $(g_1,g_2,...,g_n)$ where $g_i in G_i$. Show that the operation $bullet$: $G times G rightarrow G$ given by $(g_1,g_2,...,g_n) bullet (g'_1,g'_2,...,g'_n) = (g_1g'_1,g_2g'_2,...,g_ng'_n)$ makes $G$ into a group.




Proof of associativity: $((g_1,g_2,...,g_n) bullet (g'_1,g'_2,...g'_n)) bullet (g''_1,g''_2,...,g''_n)$ = $(g_1g'_1,g_2g'_2,...,g_ng'_n) bullet (g''_1,g''_2,...,g''_n) $ = $(g_1g'_1g''_1,g_2g'_2g''_2,...,g_ng'_ng''_n)$ = $(g_1,g_2,...,g_n) bullet (g'_1g''_1,g'_2g''_2,...,g'_ng''_n) = (g_1,g_2,...,g_n) bullet ((g'_1,g'_2,...,g'_n) bullet (g''_1,g''_2,...,g''_n))$





Proof of the existence of an identity For each $G_i in G$, take its identity element $e_i$ and multiply it as follows: $(g_1,g_2,...,g_n) bullet (e_1,e_2,...,e_3)$ = $(g_1,g_2,...,g_n)$.





Proof of the existence of an inverse: Since each $G_i in G$ is a group, simply takes its multiplicative inverse $g_i^{-1}$ and multiply it: $(g_1,g_2,...,g_n) bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.






proof-verification finite-groups direct-product







share|cite|improve this question











$endgroup$




I am trying to prove the following and am looking toward the math.stackexchange community to comment on whether I am on the right track or not. Thank you in advance.




Let $n geq 1$ be an integer. Suppose ${G_1,...,G_n}$ is a finite set of groups. Define the set $G = G_1 times G_2 times ... times G_n$ of ordered n-tuples $(g_1,g_2,...,g_n)$ where $g_i in G_i$. Show that the operation $bullet$: $G times G rightarrow G$ given by $(g_1,g_2,...,g_n) bullet (g'_1,g'_2,...,g'_n) = (g_1g'_1,g_2g'_2,...,g_ng'_n)$ makes $G$ into a group.




Proof of associativity: $((g_1,g_2,...,g_n) bullet (g'_1,g'_2,...g'_n)) bullet (g''_1,g''_2,...,g''_n)$ = $(g_1g'_1,g_2g'_2,...,g_ng'_n) bullet (g''_1,g''_2,...,g''_n) $ = $(g_1g'_1g''_1,g_2g'_2g''_2,...,g_ng'_ng''_n)$ = $(g_1,g_2,...,g_n) bullet (g'_1g''_1,g'_2g''_2,...,g'_ng''_n) = (g_1,g_2,...,g_n) bullet ((g'_1,g'_2,...,g'_n) bullet (g''_1,g''_2,...,g''_n))$





Proof of the existence of an identity For each $G_i in G$, take its identity element $e_i$ and multiply it as follows: $(g_1,g_2,...,g_n) bullet (e_1,e_2,...,e_3)$ = $(g_1,g_2,...,g_n)$.





Proof of the existence of an inverse: Since each $G_i in G$ is a group, simply takes its multiplicative inverse $g_i^{-1}$ and multiply it: $(g_1,g_2,...,g_n) bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.






proof-verification finite-groups direct-product




proof-verification finite-groups direct-product






share|cite|improve this question















share|cite|improve this question













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edited Dec 22 '18 at 4:09









Shaun

9,241113684




9,241113684










asked Feb 24 '15 at 6:07









letsmakemuffinstogetherletsmakemuffinstogether

9441717




9441717








  • 1




    $begingroup$
    Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
    $endgroup$
    – Timbuc
    Feb 24 '15 at 6:27












  • $begingroup$
    Thank you very much Timbuc for that tip on associativity.
    $endgroup$
    – letsmakemuffinstogether
    Feb 25 '15 at 3:28














  • 1




    $begingroup$
    Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
    $endgroup$
    – Timbuc
    Feb 24 '15 at 6:27












  • $begingroup$
    Thank you very much Timbuc for that tip on associativity.
    $endgroup$
    – letsmakemuffinstogether
    Feb 25 '15 at 3:28








1




1




$begingroup$
Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
$endgroup$
– Timbuc
Feb 24 '15 at 6:27






$begingroup$
Looks fine, and extremely boring as this thing usually is. Just remember to briefly explain that in the associativity part you're using the corresponding associativity of $;G_i;$ in each coordinate, and don't drop the parentheses: $$left((g_1,...)(g_1',...)right)cdot(g_1'',...)=((g_1g_1')g_1'',...)=(g_1 (g_1'cdot g_1''),...);,;;etc.$$
$endgroup$
– Timbuc
Feb 24 '15 at 6:27














$begingroup$
Thank you very much Timbuc for that tip on associativity.
$endgroup$
– letsmakemuffinstogether
Feb 25 '15 at 3:28




$begingroup$
Thank you very much Timbuc for that tip on associativity.
$endgroup$
– letsmakemuffinstogether
Feb 25 '15 at 3:28










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$begingroup$

Just as @Timbuc says, it looks fine (but one could argue that you need to include some reference to the fact that $G$ is closed under $bullet$, but that follows by construction of $bullet$).






share|cite|improve this answer









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    $begingroup$

    Just as @Timbuc says, it looks fine (but one could argue that you need to include some reference to the fact that $G$ is closed under $bullet$, but that follows by construction of $bullet$).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just as @Timbuc says, it looks fine (but one could argue that you need to include some reference to the fact that $G$ is closed under $bullet$, but that follows by construction of $bullet$).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just as @Timbuc says, it looks fine (but one could argue that you need to include some reference to the fact that $G$ is closed under $bullet$, but that follows by construction of $bullet$).






        share|cite|improve this answer









        $endgroup$



        Just as @Timbuc says, it looks fine (but one could argue that you need to include some reference to the fact that $G$ is closed under $bullet$, but that follows by construction of $bullet$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 4:07









        ShaunShaun

        9,241113684




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