There is this exercise in my abstract algebra notes:
Show that the subgroups of $\mathbb Z / 4 \mathbb Z$ are $\{\bar0\}$, $\{\bar0,\bar2\}$ and $\mathbb Z / 4 \mathbb Z$
Now I know the criteria for a subgroup
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(i) That the identity element be an element of said subgroup
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(ii) That for any two elements the result of summing them in this case will be an element
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(iii) That an inverse exists for each element
Now I'm asking if there is a way to algebraically prove this instead of checking each subgroup individually, also is it a given that for a congruence class the inverse will be an element, i. e.
$$a \in H \implies a^{-1}\in H$$
Thanks for reading any help is appreciated!
Best Answer
In general, no. In $\mathbb{Z}$, the inverse of $1$ is $-1$, but $1 \in \overline{1}$ since $4 | (1 - 1)$ and $-1 \in \overline{3}$ since $4 | (-1 - 3)$.
In terms the the congruence classes themselves, the inverse of any congruence class $\overline{a} \in \mathbb{Z}/4\mathbb{Z}$ is actually another congruence class (which may or may not be the same as $\overline{a}$).
For example, the inverse of $\overline{1}$ is $\overline{3}$ since $\overline{1} + \overline{3} = \overline{1 + 3} = \overline{4} = \overline{0}$, whereas the inverse of $\overline{2}$ is $\overline{2}$ since $\overline{2} + \overline{2} = \overline{2+2} = \overline{4} = \overline{0}$.