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?

 A: $\{1,4\}\le\Bbb Z_5^*$ because it meets the subgroup criterion:  it's closed under $x,y\mapsto xy^{-1}$.
Other combinations are $4\cdot4=16\cong1\bmod5$ and $1\cdot1\cong1\bmod5$. 
So $4$ has order two.  
A: 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 $11^{-1}=11$.
A: A subset H of a group G, with operation $*$, is a subgroup if:
0) H is closed respect the operation of G, that is: for every $a,b \in H$, holds $a*b\in H$;
1) It's satisfied the associative property.
2) The identity element of G belongs to H.
3) Every element of H has an inverse, that is: for every $a\in H$, exists $b\in H$ such that a*b = 1.
Warning: in this definition $*$ is a generic operation, it isn't necessarily the usual product of integers.
You can verify those conditions for your examples. I'll write the first to help you.
$\{0,2,4,6\}$ in $( \mathbb{Z}_8, +)$ is a subgroup because:
0) Every time we sum two elements of $\{0,2,4,6\}$ we obtain an element of $\{0,2,4,6\}$:
$0+0=0,\hspace{5mm} 0+2=0,\hspace{5mm} 0+4=4,\hspace{5mm} 0+6=0$
$2+0=2,\hspace{5mm} 2+2=4,\hspace{5mm} 2+4=6,\hspace{5mm} 2+6=8=0$
$4+0=4,\hspace{5mm} 4+2=6,\hspace{5mm} 4+4=8=0,\hspace{5mm} 4+6=10=2$
$6+0=6,\hspace{5mm} 6+2=8=0,\hspace{5mm} 6+4=10=2,\hspace{5mm} 6+6=12=4$
1) The associative property is satisfied in every subset: we never need to verify something about this.
2) $0$ is the identity element of $(\mathbb{Z}_8, +)$, and it belongs to $\{0,2,4,6\}$.
3) Every element of $\{0,2,4,6\}$ has an inverse:
$0$ has inverse $0:\hspace{2mm} 0+0=0$
$2$ has inverse $6:\hspace{2mm} 2+6=0$
$4$ has inverse $4:\hspace{2mm} 4+4=0$
$6$ has inverse $2:\hspace{2mm} 6+2=0$
You can do the same checks for the other two examples.
Just remember the definition of $\mathbb{Z}^*_m$ : it's the group formed by:
a) the subset of elements of $\mathbb{Z}_m$ that are coprime with $m$,
b) the operation is the modular product (not + as in $\mathbb{Z}_m$ !).
For examples:
$\mathbb{Z}^*_5 = \{1,2,3,4\}$
$\mathbb{Z}^*_{12} = \{1,5,7,11\}$ ... so here answer to your example is immediate (why?).
