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The ideal is defined in the ring theory;

In ring theory, a branch of abstract algebra, an ideal of a ring is a special subset of its elements

In the answer to this question What is the exponent of a group?, the term ideal is used for groups as;

The exponent of a group $G$ is the non-negative generator of the ideal $\{z \in \mathbb{Z} : \forall g \in G (g^z=1)\}$.

Is this usage of the term ideal correct? If not, what is the correct term?

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    $\begingroup$ $\{z \in \mathbb{Z} : \forall g \in G (g^z=1)\}$ is a subset of $\mathbb Z$, and $\mathbb Z$ is certainly a ring. So, the answer is yes. $\endgroup$
    – azif00
    Commented Jan 17, 2021 at 22:48

2 Answers 2

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The ideal $\{z \in \mathbb{Z} : \forall g \in G (g^z = 1)\}$ does not live in $G$, it lives in $\mathbb{Z}$. Because $\mathbb{Z}$ is a PID whose units are $\{1, -1\}$, we may speak of "the non-negative generator" of any ideal.

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This can be verified directly from the definition of an ideal.

Let $\mathbf I = \{z \in \mathbb{Z} : \forall g \in G (g^z=1)\} \subset \mathbb Z.$

Suppose $z_1, z_2 \in \mathbf I$. Then $\forall g \in G, g^{z_1} = g^{z_2} =1$. So $\forall g \in G, g^{(z_1+z_2)} = g^{z_1} g^{z_2} =1$. Hence $z_1+z_2 \in \mathbf I$.

Suppose $k \in \mathbb{Z}$ and $z \in \mathbf I$. Then $\forall g \in G,g^{z} = 1$. So $g^{kz} = (g^{z})^k =1$. Hence $kz \in \mathbf I$.

It follows that $\mathbf I$ is an ideal of $\mathbf Z$.

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