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The problem: Prove that a group of order 20 has a subgroup of order 5.

Background: This problem arises from a discrete mathematics homework assignment. Our course materials cover cosets, Lagrange's theorem, the order of group elements, and cyclic groups but not Cauchy's theorem or Sylow's theorem, so I'm looking for a proof within these limitations.

My attempts: By Lagrange, all elements have order 1, 2, 4, 5, 10 or 20. If there is an element $x$ of order 20 then $x^4$ has order 5. if there is an element $x$ of order 10 then $x^2$ has order 5. Hence, we may assume that all elements have order 1, 2 or 4. I'm stuck at going further to reach a contradiction.

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    $\begingroup$ By Lagrange all elements have order $1,2,4,5,10$ or $20$. If there is an element $x$ of order $20$ then $x^4$ has order $5$. if there is an element $x$ of order $10$ then $x^2$ has order $5$. Hence, you may assume that all elements have order $1,2$ or $4$ and find a contradiction. $\endgroup$
    – Marcos
    Commented May 30 at 10:32
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    $\begingroup$ I understand, but still Cauchy's theorem belongs to basic group theory, and not to advanced group theory. For the proof, the Sylow theorems are not necessarily needed. In this post, there is also a proof without Sylow or Cauchy, with a reference, how to see that a cyclic subgroup of order $5$ is normal, in this case. $\endgroup$ Commented May 30 at 10:53

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I think Cayley action and parity of permutations are more elementary than Cauchy or Sylow. Otherwise picking up things from the point you reached.

So let $G$ be a group of order $20$, and assume that all the elements have orders $1,2$ or $4$. If $g\in G$ has order four, then the (Cayley) permutation $\rho(g):G\to G$, $x\mapsto xg$ is a product of disjoint 4-cycles, necessarily five of them. Therefore $\rho(g)$ is an odd permutation.

But the mapping $f:z\mapsto \mathrm{sign}(\rho(z))$ from $G$ to $(\{\pm1\},\cdot)$ is a homomorphism of groups. We just saw that every element of order four is mapped to $-1$. On the other hand, if $x$ has order two, then $\rho(x)$ is a product of ten disjoint 2-cycles, and therefore $f(x)=+1$.

Assume first that there exists at least one element of order $4$. It follows from the first isomorphism theorem that the kernel of $f$ is a subgroup $H$ of order ten, and in our case $H$ consists of elements of order two only. Repeat the dose, and consider the Cayley action on $H$. Restricted to $H$ every permutation $\rho(h), h\in H$, is a product of five disjoint 2-cycles. This is impossible, because this would apply to all the elements of $H$, but the product of two odd permutations is an even permutation.

We are left with the possibility that $G$ has no elements of order $4$ either. In other words, all the non-identity elements of $G$ have order two. A well known exercise is to show that $G$ is then abelian. This leads to a contradiction as follows. Let $x,y,z\in G\setminus\{1\}$ be distinct elements such that $z\neq xy$. The subgroup $K_2$ generated by $x$ and $y$ has exactly four elements, namely $$K_2=\{1,x,y,xy\}.$$ Those need to be included, but by commutativity and the order information, this set is closed under multiplication, and hence $K_2\le G$. But, it follows that the subgroup generated by all three is $$K_3=K_2\cup K_2z.$$ This subset has $8$ elements. But, by Lagrange, $G$ cannot have a subgroup of order eight. Again, we have a contradiction, so it is impossible that all the elements of $G$ would have orders $1,2$ or $4$.

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    $\begingroup$ I suspect something simpler may be out there. Didn't find anything, and the answers to the question Dietrich linked to, don't cover this. $\endgroup$ Commented May 30 at 12:43
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    $\begingroup$ It is a bit difficult (for me at least) to gauge which argument would be the most elementary. On the courses I teach all the material I needed in the above argument are covered well before we get to Cauchy (and even group actions other than Cayley). $\endgroup$ Commented May 30 at 12:52
  • $\begingroup$ It's well-known that all finite abelian groups are CLT groups (converse to Lagrange's theorem), so you can stop there (at $G$ is abelian). I didn't quite follow your argument at the end, but it might be me. $\endgroup$
    – i can try
    Commented May 30 at 21:28
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    $\begingroup$ @softtostadaburrito Sure, but the proof of that is likely more advanced than Cauchy's. $\endgroup$ Commented May 31 at 2:43
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Playing the devil's advocate: Isn't the proof of Cauchy's Theorem even easier? If $G$ is abelian and $5\mid|G|$, then induction on $|G|$ shows that $G$ has an element of order 5. If $G$ is the smallest group with $5\mid|G|$ and no element of order 5, then for each $1\ne x\in G$ the conjugacy class $x^G$ has size $1\ne |x^G|=|G|/|C_G(x)|$ by the Orbit-Stabilizer Theorem. Hence $|C_G(x)|<|G|$, so $5\nmid|C_G(x)|$ and thus 5 divides $|x^G|$. Since $G\setminus \{1\}$ is a union of conjugacy classes, 5 divides $|G|-1$, a contradiction. This proof works if 5 is replaced by a prime $p$. It may seem paradoxical, but sometimes it is easier to prove a harder theorem.

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Since in general $\langle x\rangle\le C_G(x)\le G$, if $x$ has order $2$ or $4$, then $|C_G(x)|$ must be a multiple of $2$ or $4$, which divides $20$, and hence it can be $2$ or $4$ or $10$. But $10$ is ruled out by the following:

Lemma. A group $H$ of order $10$ has an element of order $5$.

Proof. By contradiction, let's suppose it hasn't got any. Then every nontrivial element has order $2$. Since $4,6,8\nmid 10$, the only possible class equation reads: $$10=|Z(H)|+5k$$ for some nonnegative integer $k$. As by assumption $|Z(H)|\ne 5$, and $5\nmid 9,8$, the only option is $k=0$ and hence $H$ abelian: but then, for $x,y\in H\setminus \{1\}$, $K:=\langle x\rangle \langle y\rangle $ is a subgroup of $H$ of order $4$: contradiction (Lagrange). $\space \Box$

Therefore, the class equation for $G$ reads: $$20=|Z(G)|+5(2l+m)$$ for some nonnegative integers $l,m$: contradiction, because $|Z(G)|=1,2,4$, but $5\nmid 19,18,16$.

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I think I have an argument with no prerequisites other than Lagrange's theorem and the concept of a quotient group.

Let $G$ be the group, and assume (for contradiction) that every element has order 1, 2 or 4. We can actually write down the elements:

  • $e$ has order 1.
  • Some $x_1,x_2, \ldots, x_n$ of order 2,
  • Some $y_1, y_1^{-1}, y_2, y_2^{-1}, \ldots y_{m}, y_{m}^{-1}$ of order 4.

The product $xy$ of two elements of order 2, is again an element of order 2. This is not true in general (thank you to @Jyrki Lahtonen for pointing it out). But suppose that $xy$ has order 4 (the only other option). In this case, the subgroup generated by $x,y$ has order exactly 8. Indeed, the elements are $e, x, y, xy, yx, xyx, yxy, xyxy$. The rest of the sequences of $x,y$ pair up with these (for example $yxyx$ is the inverse of $xyxy$ because if you multiply them everything cancels, but $xyxy$ is self-inverse, so they're equal). In fact, this is a standard presentation of $D_8$. Of course $G$ cannot contain a subgroup of order 8.

Therefore the set $H = \{e, x_1, \ldots, x_n\}$ is a subgroup, in fact it's a normal subgroup (since conjugation preserves order) and it's Abelian (classic exercise). This implies its order is a power of 2 (also classic exercise, independent of Cauchy's theorem), so $n=1$ or $n=3$ (not a useful fact, but it is true).

So we can form the quotient group $G/H$. What does it look like? Well it's got $H$ in it, and then every element of the form $y_{m}H$ or $y_{m}^{-1}H$. They might not all be different (in fact they must not all be different) but nevertheless, every non-identity element of $G/H$ is of the form $\text{(element of order $4$)}H$.

Well, obviously every element of order 4 squares to an element of order 2 (i.e. belonging to $H$). So $G/H$ is also a group whose every non-identity element has order 2. So its size is a power of 2 as well.

But this implies 20 is a product of two powers of 2 (which it isn't). So you're done.

Incidentally, this argument generalises to any group of order $4p$, except of course when $p=2$, in which case you can use it to classify the groups of order 8.

Classification of groups of order 8

In the case $p=2$, we can run through the argument step by step and check off groups of order 8 until we exhaust them all.

The first thing to check is the element orders. If $|G|=8$ then of course it is possible for an element to have order 8. If it does, then it's a generator, and the group is cyclic. In other words, this case is $G=C_8$ and in the remaining cases all elements have orders 1, 2 or 4 as before.

Again it is very possible that the order-2 subset is not a subgroup at all. But the argument above shows that in this case $G$ has a subset isomorphic to $D_8$, and in fact this shows $G=D_8$. In the remaining cases we have a normal subgroup.

Is it a proper subgroup? Well, not if every non-identity element has order 2. And this can happen, at least in the case $(C_2)^3$. Indeed, this is the only possible group structure, since $G$ is in this case a vector space over $\mathbb{F}_2$.

That covers the case where the subgroup has order 8. It could also have size 2 or 4. If it has size 2 then there are six elements of order 4 (so three different ones, if you account for inverse pairs). Call them euphemistically $i,j,k$. We know $i^2=j^2=k^2$ because each of these is an element of order 2, and we only have one. This implies that the unique element of order 2 commutes with all six of the order-4 elements, so we can (again euphemistically) call it $-1$, and realise we have found the quaternions.

Finally if there are 3 elements of order 2 (and therefore 4 elements of order 4), take any of the order-4 elements $y$. Of course $y^2$ is an element of order 2. If $x$ is one of the other elements of order 2 (either of them) then I claim $G \simeq \langle x \rangle \times \langle y \rangle$. Indeed, since $\langle y \rangle$ is normal in $G$ (it has index 2), this product clearly spans $G$. The two subgroups have empty intersection by construction. This is enough to prove the isomorphism.

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  • $\begingroup$ Otherwise nice, but why is $H$ a subgroup? It is possible for the product of two elements of order two to have order four. Happens for example in the dihedral group $D_4$ of symmetries of a square. $\endgroup$ Commented Jun 3 at 20:47
  • $\begingroup$ @JyrkiLahtonen Oh no! You are right of course. I'm not sure this can be saved! $\endgroup$ Commented Jun 3 at 22:07
  • $\begingroup$ @JyrkiLahtonen I was wrong, it can be rescued! Let me know if you are happy with the modified answer (and thank you for your correction). $\endgroup$ Commented Jun 4 at 19:11
  • $\begingroup$ I should have noticed that $D_8$ did not arise in the classification of my original argument... $\endgroup$ Commented Jun 4 at 19:43
  • $\begingroup$ Your fix works, I think :-) $\endgroup$ Commented Jun 5 at 16:43
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(Adapted from C. Jordan, Traité des substitutions et des équations algébriques).

Let $G$ be a group of order $20$. Embedd it into $S= S_{20}$. Consider $H < S$ of order $5^4$ ( a direct product of $4$ cyclic groups of order $5$ )

Assume that $G$ has no element of order a multiple of $5$. Then every conjugate of $G$ in $S$ intersects $H$ only in $e$. Consider the action of $G \times H$ on $S$, $(g,h)\cdot s = g s h^{-1}$. Then the stabilizer of every point is trivial, so every orbit has cardinality $|G \times H|$. We conclude that $|G|\cdot |H|$ divides $|S_{20}|$, so $5^5$ divides $20!$, contradiction.

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    $\begingroup$ Apparentlly this is Cauchy original proof (1815). Jordan published his book in 1870. Sylow 's paper appeared in 1872. $\endgroup$
    – orangeskid
    Commented Jun 3 at 21:12
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    $\begingroup$ Thanks for sharing! $\endgroup$ Commented Jun 4 at 2:41
  • $\begingroup$ @orangeskid Cauchy's original proof of what - Cauchy's theorem for arbitrary $5=p$? $\endgroup$ Commented Jun 4 at 17:53
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    $\begingroup$ @preferred_anon: Yes, for arbitrary p. $\endgroup$
    – orangeskid
    Commented Jun 4 at 19:34

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