Ideal of a group

Question

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?

Charles Hudgins · Accepted Answer

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.

Correct answer by Charles Hudgins on January 26, 2021

steven gregory · Answer

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 $mathbf Z$.

Ideal of a group

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