# Ideal of a group

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?

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

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.
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 $$\mathbf Z$$.