<s>[INST] Explain why $\{0,2,4,6\}\le\Bbb Z_8$, $\{1,4\}\le\Bbb Z_5^*$, and $\{1,5,7,11\}\le\Bbb Z_{12}^*.$ Could someone please explain the examples?
I can see that the first example is using modular addition, where 4+6=2. But I lack the maths background to understand the next two examples, I thought the * symbol mean multiply?

 [/INST] It suffices to check for closure, an identity, and inverses, since associativity is inherited from each parent group; that they are each subsets is clear. I will demonstrate these properties by means of their Cayley tables.
For $\{0,2,4,6\}\le \Bbb Z_8$, we have
$$\begin{array}{c|cccc}
+_8 & 0 & 2 & 4 & 6\\
\hline 
0 & 0 & 2 & 4 & 6 \\
2 & 2 & 4 & 6 & 0 \\
4 & 4 & 6 & 0 & 2 \\
6 & 6 & 0 & 2 & 4,
\end{array}$$
from which one can see that the identity is $0$ and $-2=6$, and $-4=4$.
For $\{1,4\}\le\Bbb Z_5^*$, we have
$$\begin{array}{c|cc}
\times_5 & 1 & 4 \\
\hline
1 & 1 & 4\\
4 & 4 & 1,
\end{array}$$
from which one can see that the identity is $1$ and $4^{-1}=4$.
For $\{1,5,7,11\}\le\Bbb Z_{12}^*$, we have
$$\begin{array}{c|cccc}
\times_{12} & 1 & 5 & 7 & 11\\
\hline 
1 & 1 & 5 & 7 & 11 \\
5 & 5 & 1 & 11 & 7 \\
7 & 7 & 11 & 1 & 5 \\
11 & 11 & 7 & 5 & 1,
\end{array}$$
from which one can see that $1$ is the identity, $5^{-1}=5, 7^{-1}=7$, and $11^{-1}=11$.
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