Classification of groups of order 20

Classify the groups of order 20. [Hint: There are five isomorphism types.]


Note that 20 = 2^2 \cdot 5.

By FTFGAG, there are two distinct abelian groups of order 20: Z_{20} and Z_{10} \times Z_2.

Now let G be a nonabelian group of order 20. By Sylow’s Theorem, n_5 = 1, so that G has a unique (hence normal) Sylow 5-subgroup H \cong Z_5. Now let K be any Sylow 2-subgroup of G. By Lagrange, we have H \cap K = 1, so that G = HK. By the recognition theorem for semidirect products, G \cong H \rtimes_\varphi K for some \varphi : K \rightarrow \mathsf{Aut}(H). Evidently, classifying the nonabelian groups of order 20 is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let H = Z_5 = \langle y \rangle. Note that \mathsf{Aut}(H) = \langle \alpha \rangle \cong Z_4; where \alpha(y) = y^2.

Let K = Z_4 = \langle x \rangle. There are four distinct homomorphisms K \rightarrow \mathsf{Aut}(H).

If \varphi_1(x) = 1, then \varphi_1 is trivial; this contradicts the nonabelianicity of G.

If \varphi_2(x) = \alpha, then Z_5 \rtimes_{\varphi_2} Z_4 is indeed a nonabelian group of order 20.

If \varphi_3(x) = \alpha^2, then Z_5 \rtimes_{\varphi_3} Z_4 is indeed a nonabelian group of order 20. Moreover, since \mathsf{ker}\ \varphi_3 \cong Z_2 and \mathsf{ker}\ \varphi_2 \cong 1, H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K.

If \varphi_4(x) = \alpha^3, then \mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2. Since Z_4 is cyclic, by a previous theorem, H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K.

Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.

Suppose now that K = Z_2^2 = \langle a \rangle \times \langle b \rangle. Again, \psi : Z_2^2 \rightarrow Z_4 is determined uniquely by \psi(a) and \psi(b), and is indeed a homomorphism provided |\psi(a)| and |\psi(b)| divide 2. We thus have \psi(a), \psi(b) \in \{ 1, \alpha^2 \}, for a total of four choices.

If \psi_1(a) = \psi_1(b) = 1, then \psi_1 = 1, contradicting the nonabelianicity of G.

If \psi_2(a) = \alpha^2 and \psi_2(b) = 1, then Z_5 \rtimes_{\psi_2} Z_2^2 is indeed a nonabelian group of order 20.

If \psi_3(a) = 1 and \psi_3(b) = \alpha^2, then \varphi_3 = \varphi_2 \circ \theta, where \theta(a) = b and \theta(b) = a. Clearly \theta is an automorphism of Z_2^2. By a lemma to a previous theorem, we have H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K.

If \psi_4(a) = \alpha^2 and \psi_4(b) = \alpha^2, then \varphi_4 = \varphi_2 \circ \theta, where \theta(a) = a and \theta(b) = ab. Clearly \theta is an automorphism of Z_2^2. By a lemma to a previous theorem, we have H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K.

Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.

In summary, the distinct groups of order 20 are as follows. We let Z_5 = \langle y \rangle, Z_4 = \langle x \rangle, and Z_2^2 = \langle a \rangle \times \langle b \rangle.

  1. Z_{20}
  2. Z_{10} \times Z_2
  3. Z_5 \rtimes_{\varphi_3} Z_4, where \varphi_3(x)(y) = y^{-1}.
  4. Z_5 \rtimes_{\varphi_2} Z_4, where \varphi_2(x)(y) = y^2
  5. Z_5 \rtimes_\psi Z_2^2, where \psi(a)(y) = y^{-1} and \psi(b)(y) = y.
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Comments

  • Emily HK  On April 17, 2012 at 10:53 pm

    Did you mean /phi sub 3 instead of /phi sub 1?

    • nbloomf  On August 21, 2012 at 10:25 am

      Yes. Thanks!

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