How to show that if $G$ is union of three proper subgroups, then there is an epimorphism from $G$ to the Klein 4-group?

I'm going through Grove's "Algebra." I'm having trouble with this one:

"If $G$ is the union of three proper subgroups, then there is an epimorphism from $G$ to the Klein 4-group."

I think the subgroups should in some sense correspond to the non-identity elements in the Klein 4-group.

Solution 1:

Suppose $G=L\cup M\cup N$ with $L$, $M$, and $N$ proper subgroups of $G$. Recall that a group cannot be a union of two proper subgroups, so for each of $L$, $M$, and $N$, there exists an element that is not in the union of the other two.

Claim 1. $L\cap M=L\cap N$.

Let $u\in L\cap M$. If $u\notin N$, then let $n\in N\setminus (L\cup M)$. Then $un\notin N$, since $n\in N$ but $u\notin N$; and $un\notin L$, because $u\in L$ but $n\notin L$; and similarly $un\notin M$. Thus, $un\notin L\cup M\cup N$, a contradiction. Thus, if $u\in L\cap M$, then $u\in N$. Thus, $L\cap M\subseteq L\cap N$. The other inclusion follows symmetrically.

Claim 2. $L\cap M=L\cap N=M\cap N = L\cap M\cap N$.

The argument from Claim 1, mutatis mutandis, establishes this.

Claim 3. If $x,y\notin L$, then $xy\in L$.

Note that $x$ and $y$ each lie in at most one of $M$ and $N$, since $M\cap N\subseteq L$. If $x\in M\setminus (L\cup N)$ and $y\in N\setminus(L\cup M)$, then $xy\notin M\cup N$, hence $xy\in L$ and we are done. So assume that $x,y\in M\setminus (L\cup N)$. Let $z\in L\setminus (M\cup N)$; then $zx\notin L\cup M$, so $zx\in N\setminus (L\cup M)$, and since $y\in M\setminus(L\cup N)$ it follows that $z(xy)\notin M\cup N$, hence $z(xy)\in L$. Since $z\in L$, then $xy\in L$, as claimed.

Claim 4. If $x,y\notin M$, then $xy\in M$. If $x,y\notin K$, then $xy\in K$.

The argument from Claim 3, mutatis mutandis, establishes this claim.

Claim 5. $L\cap M\cap N\triangleleft G$.

Let $x\in L\cap M\cap N$, and $g\in G$. If $g$ is in at least two of $L$, $M$, and $N$, then it lies in all three so $gxg^{-1}\in L\cap M\cap N$. If $g$ is in exactly one of $L$, $M$, and $N$, say $L$, then $gx\notin M$, and $g^{-1}\notin M$, hence by Claim 4 we have $gxg^{-1}\in M$; likewise, $gx\notin N$, $g^{-1}\notin N$, so $gxg^{-1}\in N$. Thus $gxg^{-1}\in M\cap N = L\cap M\cap N$, which proves that $L\cap M\cap N$ is normal.

Corollary. $G/(L\cap M\cap N)$ is isomorphic to the Klein $4$-group.

Proof. The nontrivial elements of $G/(L\cap M\cap N)$ correspond to cosets represented by elements that lie in exactly one of $L$, $M$, and $N$. If $g$ and $g'$ lie in $L$ but not in $M\cup N$, then $g(L\cap M\cap N) = g'(L\cap M\cap N)$, since $g'^{-1}g\in L\cap M\cap N$ by Claims 3 and 4. That is, we have exactly one coset corresponding to elements in $L$ but not in the triple intersection, one for $M$, and one for $K$. Thus, $G/(L\cap M\cap N)$ has exactly four elements; by Claims 3 and 4, each element is of exponent $2$, so $G/(L\cap M\cap N)$ is the Klein $4$-group. $\Box$

Also worth noting:

Proposition. A group $G$ is a union of three proper subgroups if and only if $G$ has a quotient isomorphic to the Klein $4$-group.

Proof. The necessity was proven above. Conversely, the Klein $4$-group $C_2\times C_2$ is the union of its three nontrivial proper subgroups, so by the isomorphism theorems, if $G/N\cong C_2\times C_2$, then the pullbacks of these three subgroups are proper and their union is all of $G$. $\Box$

Solution 2:

Let $G = I \cup J \cup K$. $G$ cannot be written as a union of fewer of those. For example, $G \not = I \cup J$, because then we would have some $i \in I$, $j \in J$ such that $ij$ is neither in $I$ nor $J$. I will first prove that if $g \in I \cap J$, then $g \in K$. Assume that $g \in (I \cap J) \, \backslash K$. Then since we have some element $k \in K \, \backslash \, (I \cup J)$, we would see that $gk$ cannot be in any of $I,J,K$, which is a contradiction.

Now, define $\varphi : G \to K_4$ taking $g \mapsto i$ if $g \in I \,\backslash\, (J \cup K)$, $g \mapsto j$ if $g \in J \, \backslash\,(I \cup K)$, and define $g \mapsto k$ similarly. Finally, let $\varphi$ take $g \to 1$ if $g \in I \cap J \cap K$.

All that is left is to show that $\varphi$ is indeed a homomorphism. If $g_1 \in I \, \backslash (J \cup K)$, $g_2 \in J \, \backslash (I \cup K)$, then $g_1 g_2$ is neither in $I$ nor $J$, so it must be in $K$ by $G = I \cup J \cup K$. Then $\varphi(g_1 g_2) = k = ij = \varphi(g_1)\varphi(g_2)$. If $g_1 \in I \, \backslash (J \cup K)$ and $g_2 \in I \cap J \cap K$, then it is clear that $g_1 g_2 \in I \, \backslash \,(J \cup K)$. Thus, $\varphi(g_1 g_2) = i = i(1) = \varphi(g_1)\varphi(g_2)$. The other cases follow similarly.

Solution 3:

Let $G = G_1\cup G_2\cup G_3$, and we want a surjective morphism $\phi:G\to V$ (where $V=\{1,a,b,ab\}$ is the Klein 4-group).

I find it helps me to state intuitively what I expect, as you've done, and then to make this precise in order to look for a construction.

I think the subgroups should in some sense correspond to the non-identity elements in the Klein 4-group.

Well, what does "correspond to" mean? A morphism, of course!

I think the subgroups should map surjectively to the subgroups generated by the non-identity elements in the Klein 4-group.

Now we're trying to do something concrete, namely find a morphism from each subgroup to the 2-element group which, when put together, give us a well defined map $G\to V$. In order for the composite map to be well defined, each component map must have the property $g\in G_i \cap G_j, i\neq j \implies \phi(g)=1$ because the identity 1 is the only overlap in the images of the components. A natural guess would be to define our morphism by $\phi(g)\mapsto 1$ if $g\in G_i\cap G_j, i\neq j$ and otherwise $\phi(g)\mapsto a$ if $g\in G_1$; $\phi(g)\mapsto b$ if $g\in G_2$; and similarly for $G_3$. Carl has given a nice proof that this is indeed a morphism.