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Change "strict subgroups" to "proper subgroups"
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@ -168,7 +168,7 @@ also form a group under $\cdot$.
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In the previous section we said that $\alpha$ is a generator of the $(p - 1)$-order
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In the previous section we said that $\alpha$ is a generator of the $(p - 1)$-order
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multiplicative group $\mathbb{F}_p^\times$. This group has _composite_ order, and so by
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multiplicative group $\mathbb{F}_p^\times$. This group has _composite_ order, and so by
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the Chinese remainder theorem[^chinese-remainder] it has strict subgroups. As an example
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the Chinese remainder theorem[^chinese-remainder] it has proper subgroups. As an example
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let's imagine that $p = 11$, and so $p - 1$ factors into $5 \cdot 2$. Thus, there is a
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let's imagine that $p = 11$, and so $p - 1$ factors into $5 \cdot 2$. Thus, there is a
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generator $\beta$ of the $5$-order subgroup and a generator $\gamma$ of the $2$-order
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generator $\beta$ of the $5$-order subgroup and a generator $\gamma$ of the $2$-order
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subgroup. All elements in $\mathbb{F}_p^\times$, therefore, can be written uniquely as
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subgroup. All elements in $\mathbb{F}_p^\times$, therefore, can be written uniquely as
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