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

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

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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: $(g_1,g_2,...,g_n) \bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.

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    $\begingroup$ 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
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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 $\bullet$).

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