I need to prove the group of the units of $\mathbb{Z}_3\times\mathbb{Z}_3$ is isomorphic to the Klein-4 group. But I'm really struggling to prove this. Any hints to start me off in the right direction??
[Math] Isomorphic to the Klein 4 group
abelian-groupsabstract-algebrafinite-groupsgroup-theory
Related Solutions
Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$...
It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can always group them into two pairs. What's remarkable is that there are exactly three such groupings: $$ \{1, 2, 3, 4\} = \{1, 2\} \sqcup \{3,4\} = \{1, 3\} \sqcup \{2,4\} = \{1, 4\} \sqcup \{2,3\}.$$
If I call these three groupings $\mathbf{Order}$, $\mathbf{Parity}$ and $\mathbf{5sum}$, every permutation of $\{1,2,3,4\}$ induces a permutation of these three things. For example, the $4$-cycle $(1234)$ induces the transposition $(\mathbf{Order}\ \mathbf{5sum})$. So I get a homomorphism $$ \mathfrak S(4) \to \mathfrak S(\{\mathbf{Order}, \mathbf{Parity}, \mathbf{5sum}\})\simeq \mathfrak S(3).$$ It's quite easy to show that this morphism is surjective (we already found the transposition $(\mathbf{Order}\ \mathbf{5sum})$ in its image, it's not hard to find another transposition or a $3$-cycle, and that will be enough to generate the whole of $\mathfrak S(3)$.)
Let's give a proof that the kernel is exactly $V_4$: let $\sigma$ be a nontrivial permutation in it. So for example, there are two elements $a \neq b$ such that $\sigma(a) = b$ (let's call $c$ and $d$ the two remaining elements) In that case, because $\sigma$ preserves the grouping $\{a,b\} \sqcup \{c,d\}$, it must send $b$ back to $a$. And because it preserves $\{a,c\} \sqcup \{b,d\}$ and it exchanges $a$ and $b$, it must exchange $c$ and $d$ as well. So $\sigma = (a\, b)(c\, d) \in V_4$.
Therefore, the factorisation theorem gives you an isomorphism $\mathfrak S(4)/V_4 \to \mathfrak S(3)$.
Of course, this proof is not the shortest (well, it certainly isn't the shortest to write, but if you're allowed to make a lot of pictures or to play with four actual tokens, essentially all the arguments become self-evident). But it really makes the isomorphism concrete. There's an isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$ because $4$ equals $2+2$ in $3$ different ways...
A final remark: if $n \neq 4$, the only proper normal subgroup of $\mathfrak S(n)$ is $\mathfrak A(n)$. That means that the situation I described here is pretty exceptional. If you want $\mathfrak S(n)$ to act on fewer than $n$ objects, the only interesting actions are :
- for $n$ arbitrary, $\mathfrak S(n)$ acts on two tokens with the following rule: even permutations do nothing, odd ones swap the two tokens (you have to admit this action is pretty boring).
- one exceptional case: $\mathfrak S(4)$ acts on three tokens, as we just saw.
I may be overenthusiastic, but this simple remark gives much cachet to this innocent-looking isomorphism. (Another peculiarity of the same kind is that the only interesting action of $\mathfrak S(n)$ on $n+1$ tokens is an action of $\mathfrak S(5)$ on 6 tokens, coming from another exceptional behaviour of the symmetric group.)
As @rain1 pointed out, we have a group $G=\{1,a,b,ab\}$, where $a$ and $b$ are different, commute and are not equal to $1$. Let us call $ab=ba=c$. Observe that $a \neq c$ and $b \neq c$. Now look at $a^2$. Then $a^2 \notin \{a,c\}$, so either $a^2=1$ or $a^2=b$. Symmetrically, either $b^2=1$ or $b^2=a$. So there are $4$ cases to consider, but by symmetry in $a$ and $b$ this boils down to only $2$. Firstly, $a^2=1$ and $b^2=1$, in this case $G \cong V_4$. And secondly, if $a^2=1$ and $b^2=a$, then $b^4=1$ and $G \cong C_4$. So no need of the structure theorem of abelian groups.
For groups of order $6$ you can proceed in a similar, but slightly more complicated way. Just applying elementary means. No Lagrange, no Cauchy.
Best Answer
There's really only one thing you can do: list the units of $\mathbb{Z}_3\times\mathbb{Z}_3$, see if there are four of them, and then figure out whether or not they form a cyclic group.