Find four groups of order 20 not isomorphic to each other and prove why they aren't isomorphic.

So far I thought of $\mathbb Z_{20}$, $\mathbb Z_2 \oplus\mathbb Z_{10}$, and $D_{10}$ (dihedral group), but I can't find another one. Would $U(50)$ work? I know it has order 20 and is cyclic but I'm not exactly sure how to move from there. Can someone to point me on the right direction?

  • $\begingroup$ There is one group that can perhaps best be described as a subgroup of $S_5$. Consider the 5-cycle $\alpha=(12345)$. Can you find a permutation $\beta$ with the property that $$\beta\alpha\beta^{-1}=\alpha^2?$$ Then take a look at the subgroup generated by $\alpha$ and $\beta$. Squaring in a cyclic group of order five is an automorphism of order four, so... $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 7:27
  • $\begingroup$ All cyclic groups of order $n$ are isomorphic to $\mathbb{Z}_n$. Edit: is $U(50)$ cyclic? It might be, but I don't see why right away. $\endgroup$ – NotAwake Nov 23 '13 at 7:27
  • $\begingroup$ Assuming your $U(n)=\Bbb{Z}_n^*$, then it is cyclic, iff $n=2,4, p^k$ or $2p^k$ for an odd prime $p$. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 7:29
  • $\begingroup$ The element 3 of U(50) has an order of 20 so it is cyclic. $\endgroup$ – MathematicalAnomaly Nov 23 '13 at 7:29
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    $\begingroup$ Or in yet other words. This group is not on the list of "standard" groups encountered in the first course on groups. Finding the fourth is meant to be a tough exercise. I suggested looking inside $S_5$, because A) you surely know what $S_5$ is, B) it is the simplest "standard" group that has the missing group as a subgroup. GO FIND IT! $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 8:00

Note that $20 = 2^2 \cdot 5$.

By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.

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 $H \cong \mathbb{Z}_5$. 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 $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.

Let $K = \mathbb{Z}_4 = \langle x \rangle$. 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 $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.

If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ 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 $\mathbb{Z}_4$ 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 $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ 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 $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ 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 $\mathbb{Z}_2^2$. 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 $\mathbb{Z}_2^2$. 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 $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.


$Z_{10} \times Z_2$,

$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.

$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$

$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.

(Source: Crazyproject)

  • $\begingroup$ +1 for a fine solution. I'm not sure that a beginner is well placed to fully appreciate this, but it is still a fine answer. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 7:54
  • $\begingroup$ Thank you! Because I just thought he just wanted to find 4 non-isomorphic groups of oder 20 in any way. So sorry :D $\endgroup$ – An Khuong Doan Nov 23 '13 at 7:56
  • $\begingroup$ It's a good answer, but I agree that it seems beyond the level of his course. Though I wonder how the professor expects students to find one of the other two possibilities if they only just learned direct sums... $\endgroup$ – NotAwake Nov 23 '13 at 7:59
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    $\begingroup$ @azimut: $H$ is here the unique subgroup of its order. And it's abelian, so its centralizer is $$H\rtimes \ker(\varphi)\cong H\times \ker(\varphi).$$ Also $\ker(\varphi)$ is a 2-group, so we can identify it from knowing the centralizer of the Sylow 5-subgroup of $G$. Probably there are other ways. $\endgroup$ – Jyrki Lahtonen Nov 26 '13 at 21:44
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    $\begingroup$ @user72694: The group I had in mind (#4 in SnowAngel's list) does not have a faithful representation of dimension $<4$. Its irreducible characters have dimensions 1,1,1,1 and 4. It is the subgroup of permutations of vertices of the graph $K_5$ that preserves the partition of edges to the pentagon and the pentagram: $D_5$ maps both the pentagon and pentagram to itseld, but this group includes permutations that map the pentagon to pentagram and vice versa. It is also the Galois group of the polynomial $x^5-2$. $\endgroup$ – Jyrki Lahtonen Nov 27 '13 at 6:36

In addition to the two cyclic groups and the dihedral group $D_{10}$, consider the group $G$ of symmetries of an antiprism over the regular pentagon. It is transitive on the 10 vertices and the stabilizer of a vertex $p$ contains a nontrivial symmetry given by the reflection in the vertical plane passing through $p$ and the origin; for a total of 20 elements. This is different from $D_{10}$ because the central element (reflection in the origin in $\mathbb{R}^3$) has a 5th root in $G$, which is not the case for $D_{10}=D_5\times\mathbb{Z}_2$. One can think of $G$ as the group of symmetries of the icosahedron preserving a pair of opposite points.

  • $\begingroup$ Let's see, viewed as the group of symmetries of the $10$-gon in the plane, we have a central involution given by $diag(-1,-1)$ and also any orientation-reversing reflection. So the 2-sylow subgroup seems to be $\mathbb{Z}_2^2$. Similarly the involution in $D_5$ and the generator of $\mathbb{Z}_2$ form a 2-Sylow subgroup isomorphic to $\mathbb{Z}_2^2$. Where is the element of order 4? $\endgroup$ – Mikhail Katz Nov 29 '13 at 9:31
  • $\begingroup$ You are right, I am not sure how I managed to convince myself that $D_{10}$ had an element of order $4$. $\endgroup$ – Tobias Kildetoft Nov 29 '13 at 9:37
  • $\begingroup$ OK, no problem. Note that @Jyrki Lahtonen mentioned above that one of the two "difficult" order 20 groups does not admit any (reducible or irreducible) 3-dimensional representation. It would be interesting to identify the one I described with one of the groups on user SnowAngel6147's list. Presumably the remaining one will not have a 3-dimensional representation. $\endgroup$ – Mikhail Katz Nov 29 '13 at 9:40
  • $\begingroup$ Now I am trying to figure out if the one you mention here and the one I just wrote an answer with (the affine transformations) are actually the same. $\endgroup$ – Tobias Kildetoft Nov 29 '13 at 9:42
  • $\begingroup$ But according to the answer by Snowangel there is only one non-abelian group of order $20$ with elementary abelian $2$-Sylow subgroup. $\endgroup$ – Tobias Kildetoft Nov 29 '13 at 9:46

To supplement the great answer of SnowAngel6147, here is an alternative description of one of the non-abelian groups of order $20$:

One of the ones with a cyclic $2$-Sylow subgroup can be described as follows: Let $F$ be the field with $5$ elements, and let $G$ be the group consisting of all maps $f: F\to F$ of the form $f(x) = ax + b$ for some $a\in F\setminus\{0\}$ and some $b\in F$. This is a group with the operation of composition of functions (it is a nice exercise to check that this is in fact the case). It also clearly has order $20$, so we just need to check that it is not isomorphic to the others on the list.

Since $G$ is not abelian, we can rule out $G$ being isomorphic to one of the abelian groups of order $20$, and since the set of maps with $b=0$ is a subgroup of order $4$, this rules out being isomorphic to $D_{10}$.


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