Exercise about finding group isomorphisms So, i've just learned about groups homomorphisms and isomorphisms. I know that an homormorphism betweet two groups is a function such that 
$$\phi(a\square b) = \phi(a)\star \phi(b)$$
And when the homomosphism is bijective, then we have an isomorphism betweet the two groups.
The problem is that the exercise asks me to tell is these groups are isomorphic:
$$(\Bbb Z_6, +),  (\Bbb Z_7^*,.)$$
In this case, I'm looking at the groups:
$$(\Bbb Z_6, +) = \{0,1,2,3,4,5\}, (\Bbb Z_7^*,.) = \{1,2,3,4,5,6\}$$
However, there's no explanation of which is the homomorphism that I have to check, so I'm lost.
The same for these two groups: 
$$(\Bbb Z_2 \times\Bbb Z_2), \Bbb Z_4$$
Could somebody help me?
 A: Any homomorphism $\phi$ out of $\Bbb Z_6$ is totally dictated by $\phi(1)$, since:
$\phi(n) = n\cdot \phi(1) = \phi(1+1+\cdots+1)$ ($n$ times)
$= \phi(1) \ast \phi(1) \ast \cdots \ast \phi(1) = \phi(1)^n$ (since $\phi$ is a homomorphism, where $\ast$ is the operation in the target group).
Now for $\phi: \Bbb Z_6 \to (\Bbb Z_7)^{\times}$, this gives us $6$ choices for $\phi(1)$ (note the operation in the latter group is multiplication mod $7$).
If $\phi(1) = 1$, then $\phi(n) = 1$, for all $n \in \Bbb Z_6$, this is the trivial homomorphism, which is definitely not bijective.
This leaves us with $5$ other choices.
If $\phi(1) = -1 = 6$, we have $\phi(2) = \phi( 1+ 1) = \phi(1)\phi(1) = 6\cdot 6 = 1$ (since $36 = 1$ (mod $7$)).
This implies $\phi(4) = \phi(2 + 2) = \phi(2)\phi(2) = 1\cdot 1 = 1$, so this homomorphism is not injective either ($2 \neq 4$ in $\Bbb Z_6$, but $\phi(2) = \phi (4)$).
Down to $4$ possible choices, now.
If $\phi(1) = 2$, we have $\phi(2) = \phi(1+1) = \phi(1)\phi(1) = 2\cdot 2 = 4$. Thus $\phi(3) = \phi(2 + 1) = \phi(2)\phi(1) = 4\cdot 2 = 1$, so $\phi$ sends $0 \to 1$ and $3 \to 1$, and again it is not injective.
A similar argument shows $\phi(1) = 4$ is also not injective (the details of which I leave to you).
So let's examine $\phi(1) = 3$ (the case $\phi(1) = 5$ is similar):
$\phi(0) = 1\\ \phi(1) = 3\\ \phi(2) = 3\cdot 3 = 2\\ \phi(3) = 3\cdot 3\cdot 3 = 2\cdot 3 = 6\\ \phi(4) = 3\cdot 3\cdot 3\cdot 3 = 6  \cdot 3 = 4\\ \phi(5) = 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 = 4\cdot 3 = 5$
This is clearly a bijection, and we have (by the way we define $\phi$) that:
$\phi(k+m) = \phi((k+m)\cdot 1) = [\phi(1)]^{k+m} = \phi(1)^k\phi(1)^m = \phi(k\cdot 1)\phi(m\cdot 1) = \phi(k)\phi(m)$,
so it is an homomorphism, as well.
A: For $\Bbb{Z}_6$ and $\Bbb{Z}_7^*$, a hint: both are cyclic; can you use this to construct an isomorphism?
For groups of very small orders, you could also look at their multiplication table and see if you can identify one element with another.
The easiest way to see that two groups are not isomorphic is to find a group property that  one group has and the other doesn't have. Some of properties that are preserved by homomorphisms are the order of the groups, whether they are cyclic, whether they are abelian, and how many elements of a certain order they have. 
