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Exercise 16/chapter 3 from Lang's Algebra reads as follows: Prove that the inverse limit of a system of simple groups in which the homomorphisms are surjective is either the trivial group or a simple group.

So far i have observed the following: Let $I$ be the index set of this family. In general $I$ is partially ordered. This ordering induces an equivalence relation in $I$. The transition homomorphisms between groups whose indices belong to the same equivalence class are surjective by assumption and injective because the groups are simple. So they are isomorphisms. Also, the natural projections from the inverse limit of the family to the groups are surjective. If $I$ is totally ordered then they are injective as well and all the groups are isomorphic with each other and with the inverse limit.

The difficulty that i have is in dealing with the partial order. Any hints?

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What is the "equivalence relation on $I$" that the ordering induces? –  Arturo Magidin Nov 6 '11 at 23:48
    
If you have an inverses limit of a system of groups, then the index set is more than just a partially ordered set. Depending on your definition of inverse limit, it is either a directed partially ordered set (for every $i,j\in I$ there exists $k\in I$ such that $i\leq k$ and $j\leq k$), or an inversely directed partially ordered set (as above, but the property of $k$ is $k\leq i$ and $k\leq j$). Show that if any of the groups is trivial then the limit is trivial, and if they are all isomorphic then the simple group itself has the appropriate universal property. –  Arturo Magidin Nov 6 '11 at 23:57
    
You are a master...(The equivalence relation would be: $i$ is equivalent to $j$ if $i \le j$ or $j \le i$.) –  Manos Nov 7 '11 at 0:19
    
I.e., "comparable"? That is not an equivalence relation in an arbitrary partially ordered set. Take $P=\{1,2,3\}$, and $\leq = \{(1,1), (1,3), (2,2), (2,3), (3,3)\}$. Then $1$ and $2$ are each comparable to $3$, but not to each other. So the relation is not transitive. You would need to take the transitive closure; and in the case of a directed set, there is only one equivalence class. (And why does this matter, in any case?) –  Arturo Magidin Nov 7 '11 at 0:20
    
I suppose it does not matter... –  Manos Nov 7 '11 at 1:24

1 Answer 1

up vote 2 down vote accepted

Okay, this is getting too complicated for comments.

Let $\{S_i; (f_{ji})\}_{i\in I}$ be the inverse system. I take $I$ to be a directed partially ordered set, $S_i$ simple for each $i$, and whenever $i\leq j$, $f_{ji}$ is an onto homomorphism $f_{ji}\colon S_j\to S_i$.

Since an onto homomorphism with domain a simple group must be either the trivial map or an isomorphism, we see that for every $i$ and $j$ in $I$, with $i\leq j$, either $S_i\cong S_j$ or else $S_i=\{1\}$.

In particular, let $i$ and $j$ be arbitrary elements of $I$. Then there exists $k$ such that $i\leq k$ and $j\leq k$. Hence if $S_i$ and $S_j$ are both nontrivial, then $S_k\cong S_i$, $S_k\cong S_j$, and hence $S_i\cong S_j$. So any two nontrivial groups in our family are necessarily isomorphic.

If all groups are trivial, then the inverse limit is trivial. If not every group is trivial, then let $S$ be a simple group such that every nontrivial $S_i$ is isomorphic to $S$. I claim that in this case, the inverse limit is isomorphic to $S$.

The inverse limit is given by the subgroup of $\prod_{i\in I}S_i$ of all elements $(x_i)$ such that if $i\leq j$, the $f_{ji}(x_j) = x_i$. We may omit any component where $S_i=\{1\}$, and so we may assume that every $S_i$ is isomorphic to $S$ and that all structural morphisms are isomorphisms. But then I claim that $\lim\limits_{\leftarrow}S_i\cong S$. Indeed, for any $i$ and any $x_i\in S_i$, for every $j\geq i$ there is a unique $x_j\in S_j$ such that $f_{ji}(x_j) = x_i$. if $j\leq k$, then $f_{ji}(x_j) = x_i = f_{ki}(x_k) = f_{ji}(f_{kj}(x_k))$, so the uniqueness guarantees that $f_{kj}(x_k) = x_j$. We can complete the terms "downstream" as well: given any $j$, find $k$ with $i,j\leq k$; then take $x_j = f_{kj}(x_k)$, where $x_k$ was defined as above. So we get a consistent family that has $x_i$ in the $i$th coordinate. Hence $\pi_i$, the projection onto the $i$th coordinate, is onto.

I claim that it is one-to-one; indeed, if $\pi_i(x_i) = 1$, then $x_i=1$. Now let $j\in I$ be arbitary. Then there exists $k\in I$ with $i,j\leq k$; since $f_{ki}$ is an isomorphism and $f_{ki}(x_k) = x_i = 1$, then $x_k=1$; hence $f_{kj}(x_k) = x_j = 1$ as well. Thus, $(x_i)= 1$, so $\pi_i$ is one-to-one. Since $\pi_i$ is one-to-one and onto, $\pi_i$ is an isomorphism, so $\lim\limits_{\leftarrow}S_i \cong S_i \cong S$, as desired.

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