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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.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles May 24 '12 at 0:14
    
What have you tried? Can you prove either direction? –  Chris Eagle May 24 '12 at 0:18
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@ZevChonoles: Sorry about that. I have edited my original post. –  Steven Li May 24 '12 at 0:20
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There are some hints at groupprops.subwiki.org/wiki/… –  Gerry Myerson May 24 '12 at 1:27
    
This has surely been asked before on this site. –  Mariano Suárez-Alvarez May 24 '12 at 2:02

3 Answers 3

up vote 8 down vote accepted

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 $G$. $\Box$

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You wrote a very nice answer a while ago about this on MO, if I recall correctly, no? –  Mariano Suárez-Alvarez May 24 '12 at 2:37
    
There's one missing sign of dollars in the 3rd line of Claim 5 which messes up the last two lines. I tried twice to edit it and add that "$" but I keep getting a notification that an "edit" must contain at least 6 characters... –  DonAntonio May 24 '12 at 2:43
    
@DonAntonio,you could always add a «Thank you» as padding :D –  Mariano Suárez-Alvarez May 24 '12 at 2:55
    
@MarianoSuárez-Alvarez: I think you are talking about this one, though it doesn't include any proofs. –  Arturo Magidin May 24 '12 at 2:56
    
@ArturoMagidin: Thanks! –  Steven Li May 24 '12 at 21:34

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.

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Thanks guys for the comments earlier. Sorry, I got freaked out for a second and deleted it before realizing what my mistake was. @Gerry it was supposed to be $\cap$. –  Carl May 24 '12 at 1:52
    
Why didn't you just edit an undelete your previous answer? –  Arturo Magidin May 24 '12 at 1:57
    
It disappeared from the page after I had deleted it for some reason. –  Carl May 24 '12 at 2:00

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.

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