Equivalent characterization of group congruences

Let G be a group. Assume that \mathcal{P} = \{ A_i \ |\ i \in I \} is a partition of G such that \mathcal{P} is a group under the “quotient” operation as follows: to compute the product of A_i and A_j take any representatives a_i \in A_i and a_j \in A_j and let A_iA_j be the element of \mathcal{P} containing a_ia_j. (Assume this operation is well defined.)

Prove that the element N of \mathcal{P} containing 1 is a normal subgroup of G and that the elements of \mathcal{P} are the cosets of N, so that \mathcal{P} = G/N.


We will denote the \mathcal{P} equivalence class of x by [x]; hence N = [1].

Lemma 1: Let G and \mathcal{P} be as described above. Then for all g,x \in G, g[x]g^{-1} = [gxg^{-1}]. Proof: (\subseteq) we have g[x]g^{-1} \subseteq [g][x][g^{-1}] = [gxg^{-1}]. (\supseteq) We have g^{-1}[gxg^{-1}]g \subseteq [g^{-1}gxg^{-1}g] = [x], so that [gxg^{-1}] \subseteq g[x]g^{-1}. \square

Lemma 2: Let G and \mathcal{P} be as described above. Then for all g,x \in G, g[x] = [gx]. Proof: (\subseteq) We have g[x] \subseteq [g][x] = [gx]. (\supseteq) We have g^{-1}[gx] \subseteq [g^{-1}gx] = [x], so that [gx] \subseteq g[x]. \square

First we show that N \leq G. Note that N is nonempty since 1 \in N. Now suppose x,y \in N. We have 1 = [x^{-1}x] = [x^{-1}][x], and so by the uniqueness of group inverses, [x]^{-1} = [x^{-1}]. Thus [xy^{-1}] = [x][y]^{-1} = [1][1]^{-1} = [1] = N, and so xy^{-1} \in N. By the Subgroup Criterion, N is a subgroup of G.

Now we show that N is normal in G. Let g \in G; now gNg^{-1} = g[1]g^{-1} = [gg^{-1}] = [1] = N by Lemma 1.

Now we show that every A \in \mathcal{P} is a coset of N. Let A \in \mathcal{P} and a \in A. Now A = [a] = a[1] = aN, using Lemma 2. So A is a coset of N.

Now we show that every coset of N is in \mathcal{P}. Let g \in G. Since \mathcal{P} is a partition, g \in A for some A \in \mathcal{P}. Thus A = [g] = g[1] = gN, using Lemma 2, and so gN \in \mathcal{P}.

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Comments

  • Gobi Ree  On November 22, 2011 at 8:50 pm

    I like your notations. A typo: xY^{-1} should be xy^{-1}.

    • nbloomf  On November 22, 2011 at 11:51 pm

      Fixed. Thanks!

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