15

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?

MathematicalAnomaly
  • 1,021
  • 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... – Jyrki Lahtonen Nov 23 '13 at 07:27
  • 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. – NotAwake Nov 23 '13 at 07:27
  • Assuming your $U(n)=\Bbb{Z}_n^*$, then it is cyclic, iff $n=2,4, p^k$ or $2p^k$ for an odd prime $p$. – Jyrki Lahtonen Nov 23 '13 at 07:29
  • The element 3 of U(50) has an order of 20 so it is cyclic. – MathematicalAnomaly Nov 23 '13 at 07:29
  • Right, $50$ has a primitive root. – NotAwake Nov 23 '13 at 07:31
  • Or if you don't want to play the permutation game, then a semidirect product is the way to go. Hint#1: You have found all the abelian ones - stop looking. Hint#2: If a group of order 20 has an element of order 10, then it is either abelian or $D_{10}$. The missing group does not have elements of order ten. – Jyrki Lahtonen Nov 23 '13 at 07:35
  • Are you telling me a fourth group doesn't exist? You said that I found all the abelian ones and that it is either abelian or D10 wich I already have. – MathematicalAnomaly Nov 23 '13 at 07:41
  • There is a non-abelian group of order $20$ with no element of order $10$. (If it has an element of order $10$ and is non-abelian, then you can conclude its $D_{10}$.) What level is the course that this is for? – NotAwake Nov 23 '13 at 07:42
  • Thank you this makes sense. I can't fathom why my professor gave us this question as a quiz if it has no answer. – MathematicalAnomaly Nov 23 '13 at 07:44
  • It has an answer. Did you try Jyrki's hint? – NotAwake Nov 23 '13 at 07:46
  • It's undergraduate level abstract algebra We just finished learning about direct sums and this was our class quiz for the chapter. – MathematicalAnomaly Nov 23 '13 at 07:50
  • 2
    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! – Jyrki Lahtonen Nov 23 '13 at 08:00
  • Oops. I seem to indicate that there would be only 4 non-isomorphic ones. That is not correct. See SnowAngel's answer. Also what I said about non-abelian groups with an element of order 10 is not correct. The third group in SnowAngel's list shows that, and somehow I managed to ignore that group. Sorry. – Jyrki Lahtonen Nov 27 '13 at 06:23

3 Answers3

21

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 $\psi_4 = \psi_2 \circ \theta$, 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 $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.

$Z_{20}$,

$Z_{10} \times Z_2$,

$Z_5 \rtimes_{\varphi_3} Z_4$, 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)

fcm
  • 163
An Khuong Doan
  • 2,107
  • +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. – Jyrki Lahtonen Nov 23 '13 at 07:54
  • Thank you! Because I just thought he just wanted to find 4 non-isomorphic groups of oder 20 in any way. So sorry :D – An Khuong Doan Nov 23 '13 at 07:56
  • 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... – NotAwake Nov 23 '13 at 07:59
  • @NotAwake: I am wondering about the same thing. A possibility (a long shot, but I've done things like that myself as a teacher!) is that the permutation equation $\beta\alpha=\alpha^2\beta$ in $S_5$ was given as an exercise, when checking the students' mastery of algebra of permutations. I would only assign this problem by leading up to it with other problems. Without pointers like that this is a cruel one. – Jyrki Lahtonen Nov 23 '13 at 08:03
  • 1
    How exactly does $\ker(\varphi_2)\not\cong\ker(\varphi_3)$ imply $H\rtimes_{\varphi_2} K\not\cong H\rtimes_{\varphi_3} K$? – azimut Nov 26 '13 at 21:36
  • 2
    @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. – Jyrki Lahtonen Nov 26 '13 at 21:44
  • @JyrkiLahtonen, it would be interesting to see if one of the two additional groups can be represented in O(3) and if so if it can be represented by a group of symmetries of a suitable solid. – Mikhail Katz Nov 27 '13 at 05:57
  • 1
    @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 $x^5-2$. – Jyrki Lahtonen Nov 27 '13 at 06:36
  • @user72694: And groups of order 20 don't have irreducible 3D representations, because the dimension of an irreducible rep is a factor of the order of the group. – Jyrki Lahtonen Nov 27 '13 at 06:37
  • @JyrkiLahtonen, there is a order-20 group with a reducible 3-dimensional representation (the irreducible components aren't faithful) given by the symmetries of an antiprism on a regular pentagon. I originally posted it as an answer but haven't yet found an elementary argument to show that it is different from $D_{10}$ though this seems likely. – Mikhail Katz Nov 27 '13 at 13:55
  • @SnowAngel6147, we seem to have identified two nonisomorphic nonabelian groups of order 20 with 2-Sylow subgroup $Z_2\times Z_2$. How can this be reconciled with your answer? – Mikhail Katz Dec 09 '13 at 14:51
  • 2
    This answer is taken word by word from here. (Btw, nobody asked what "lemma to a previous theorem" can mean here.) –  Jan 08 '14 at 09:50
  • What is this previous lemma you refer to? – Shreya Feb 20 '19 at 21:26
2

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.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • 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? – Mikhail Katz Nov 29 '13 at 09:31
  • You are right, I am not sure how I managed to convince myself that $D_{10}$ had an element of order $4$. – Tobias Kildetoft Nov 29 '13 at 09:37
  • 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. – Mikhail Katz Nov 29 '13 at 09:40
  • 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. – Tobias Kildetoft Nov 29 '13 at 09:42
  • But according to the answer by Snowangel there is only one non-abelian group of order $20$ with elementary abelian $2$-Sylow subgroup. – Tobias Kildetoft Nov 29 '13 at 09:46
  • Are you sure? @Snow did not say this explicitly. My group has an involution given by the antipodal map, and also an involution given by a reflection in a vertical plane as described in the answer (reflection in the horizontal plane does not preserve the antiprism). So according to your interpretation of Snow, this should be isomorphic to $D_{10}$. But what about my argument with 5th root? – Mikhail Katz Nov 29 '13 at 09:58
  • Copied directly from Snow's answer (near the end) "Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup." I will need to think a bit about this. – Tobias Kildetoft Nov 29 '13 at 10:01
  • @TobiasKildetoft, I don't know what "elementary" means in this context. – Mikhail Katz Dec 01 '13 at 18:25
  • Elementary abelian means that all non-identity elements have the same order (and that the group is abelian). – Tobias Kildetoft Dec 02 '13 at 18:28
  • @TobiasKildetoft, I see. If the group of symmetries of the antiprism is isomorphic to the dihedral group, then which element corresponds to the antipodal map of the antiprism? – Mikhail Katz Dec 02 '13 at 18:32
  • One way of checking @Snow's results would be to compute the 1-dimensional cohomology group classifying these semidirect products. This won't identify the groups but it will give the right number of such groups. – Mikhail Katz Dec 02 '13 at 18:38
1

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 $D_{10}$.

Tobias Kildetoft
  • 21,247