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If $|G|=2n$ where $n$ is odd and $H$ is a subgroup of order $n$. Then $\underset{g\in G}{\prod}g\notin H$

pf: Since $G$ has even order, it must have a element of order $2$, then $\underset{g\in G}{\prod}g$ has even order. So $\underset{g\in G}{\prod}g\notin H$ as the order of everything in $H$ must divide $n$. Is this correct?

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    $\begingroup$ Is $G$ abelian? Otherwise it seems like you need to take more care in defining $\prod_{g\in G}g$. $\endgroup$ Sep 18, 2017 at 18:58
  • $\begingroup$ This is exactly how it was written $\endgroup$
    – tmpys
    Sep 18, 2017 at 19:08
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    $\begingroup$ I agree it isn't a well defined question, but the result is true for any ordering of $\prod_{g\in G}g$. $\endgroup$
    – Dap
    Sep 18, 2017 at 19:11
  • $\begingroup$ We can say something on all products $\prod_{g\in G} g$. For abelian groups this is trivial, for nonabelian groups see here. $\endgroup$ Sep 18, 2017 at 19:28

3 Answers 3

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Lemma Put $x=\prod_{g \in G} g$ and let $\sigma$ be a bijection (not necessarily an automorphism) of the set $G$. Write $x_{\sigma}=\prod_{g \in G}\sigma(g)$. Then $x \notin H$ if and only if $x_{\sigma} \notin H$.

Proof Since $|G:H|=2$, $H$ is normal and $G/H$ is abelian. Hence in $G/H$, where the order of multiplication does not matter, we have $\bar{x}=\prod_{g \in G} \bar{g}=\prod_{{g \in G,g^2=1}} \bar{g}=\prod_{g \in G} \overline{\sigma(g)}$, since the odd order elements of $G$ must lie in $H$ and do no longer appear in the product when modding out. Hence $\bar{x}=\overline{x_{\sigma}}$, that is $x=x_{\sigma}h$ for some $h \in H$ and the Lemma follows.

Now let us proceed with the proof of the question above. Let $t \in G$ be an element of order $2$ (Cauchy's Thereom assures its existence!), put $P=\{1,t\}$. Note that $t \notin H$, $G=HP$ and $H \cap P=1$. If $H=\{1, h_2, h_3, \dots, h_n\}$, we can write $G=\{1,h_2,h_3, \dots, h_{n}, t, th_2, \dots, th_n\}=\{1,h_2^{-1},h_3^{-1}, \dots, h_{n}^{-1}, t, th_2, \dots, th_n\}$. Now take the product in the last set of all the elements of $G$ as follows: $(t \cdot 1) \cdot (th_2 \cdot h_2^{-1}) \cdots (th_n \cdot h_n^{-1})=t^n=t$, since $n$ is odd and $t$ has order $2$. Because $t \notin H$, we are done by the Lemma.

Note Surprisingly, in general one can show that if $G$ is a finite group, $P \in Syl_2(G)$ and $P$ is non-cyclic or trivial (that is, $|G|$ is odd), then the set of all possible products of elements of $G$ (each taken exactly once of course) equals $G'$, the commutator subgroup. If $P$ is cyclic, then this set equals the coset $tG'$ where $t$ is the unique element of order $2$ of $P$.

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  • $\begingroup$ How do I know the odd order elements are in $H$ and what is going on with the $g^2=1$ $\endgroup$
    – tmpys
    Aug 8, 2018 at 0:07
  • $\begingroup$ Can you just give a little more justification for the chain of equalities starting with $\bar{x}$ $\endgroup$
    – tmpys
    Aug 8, 2018 at 0:08
  • $\begingroup$ The idea is to mod out to an abelian group where the order of the elements is irrelevant. Since in this case the abelian quotient group $G/H$ has order $2$, the odd order elements must lie in $H$ (Why? In general: if $f: X \rightarrow Y$ is a homomorphism of finite groups, then for each $x \in X$, $order(f(x)) \mid gcd(order(x),|Y|)$). Hence the evaluation of the product in the quotient is only relevant for the elements of $G$ of order $2$. Hope this helps. $\endgroup$ Aug 8, 2018 at 7:18
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That answer is not a valid proof. The product of the elements does not need to have the order related to one element in the product. For example, you can multiply two elements with order 2 to get an element of order 3 or 5 or any number.

Hint: This fact I think must have come up before this exercise: if a subgroup H that has half as many elements as G, then H is a normal subgroup of G.

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  • $\begingroup$ Yes, I was trying stuff with H normal, but I was not sure what to do. $\endgroup$
    – tmpys
    Sep 18, 2017 at 19:20
  • $\begingroup$ can you elaborate $\endgroup$
    – tmpys
    Sep 18, 2017 at 23:01
  • $\begingroup$ What can you do with normal subgroups? $\endgroup$ Sep 19, 2017 at 16:07
  • $\begingroup$ I never figured this out, can you help? I know if H is size n, then H is normal and G/H has order 2. Pick g in G such that g+H not equal to H. I should be able to use the fact that the product is in G to find some contradiction, but I can't figure it out. I assure you this is not for class, can you just show me what I am not thinking about. It is driving me nuts. $\endgroup$
    – tmpys
    Jul 19, 2018 at 5:43
  • $\begingroup$ "That answer is not a valid proof", which answer do you mean? Also, "The product of the elements does not need to have the order related to one element in the product": the product of all the elements of the group is again an element of the group (since $G$ is closed under the group multiplication). Hence the product of all elements (each taken once) must be one of the group elements. $\endgroup$ Jul 26, 2018 at 7:18
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Given the writing of the product, one can argue that the group actually is commutative (otherwise it would not be well-defined: the product would depend on the order in which the elements are multiplied). Hence, I do not have to care about normality, as everything is automatic.

Let $\pi$ denote the canonical surjective group homomorphism $G\rightarrow G/H$, that is the projection of $G$ onto the quotient $G/H$. This quotient is a group that contains two elements, so it is isomorphic to $\{-1, 1\}$.

What we need to show is that the element $\underset{g\in G}{\prod}g$ is not $1$ in the quotient $G/H$. So we need to compute its image by $\pi$ and check that it not $1$.

Given that $\pi$ is a group homomorphism, it is clear that

$$\pi\left(\underset{g\in G}{\prod}g\right)=\underset{g\in G}{\prod}\pi(g)$$

To compute this product, we may separate the terms whether their value is $1$ or $-1$ under $\pi$ (this is because $\pi$ is surjective, so that $G=\pi^{-1}\{1\}\sqcup\pi^{-1}\{-1\}$ as sets).

$$\underset{g\in G}{\prod}\pi(g)=\underset{g\in \pi^{-1}\{1\}}{\prod}\pi(g)\times\underset{g\in \pi^{-1}\{-1\}}{\prod}\pi(g)=\underset{g\in \pi^{-1}\{1\}}{\prod}1\times\underset{g\in \pi^{-1}\{-1\}}{\prod}-1=(-1)^{|\pi^{-1}\{-1\}|}$$

So now, all we need to do is to understand what $|\pi^{-1}\{-1\}|$ is (the cardinality of $\pi^{-1}\{-1\}$). Actually, this is $n$. Indeed, it is a known result that every cosets of a subgroup have the same cardinality, that is precisely the cardinality of the said subgroup.

So we may now conclude

$$\pi\left(\underset{g\in G}{\prod}g\right)=(-1)^{|\pi^{-1}\{-1\}|}=(-1)^n=-1$$

because $n$ is odd.

NB: Note that if $G$ is not abelian (and we assume that the product is defined after an implicit choice of the order), the above proof works just as well because the quotient, $G/H$, which is isomorphic to $\{1,-1\}$, is abelian. Hence, the manipulations I did using $\pi$, in particular the separation of the product into two parts, still holds.

One needs to justify however that $H$ is a normal subgroup of $G$. This is a result that is true for any subgroup of index $2$. A proof may be found here.

NB2: to justify in general that every cosets of a subgroup have the same cardinality, one can use the pretty straightforward argument. Say, $G$ is a finite group with subgroup $H$, and consider $gH$ a left coset of $H$. Then we have a map $H\rightarrow gH$ that sends an element $h\in H$ to $gh$. This map is bijective because it has an inverse, that is the map $gH\rightarrow H$ which sends an element $h'\in gH$ to $g^{-1}h'$.

One can adapt this easily to the case of right cosets.

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  • $\begingroup$ The group does not have to be abelian, that is the whole point. The statement is true for instance for $G=D_n$, the dihedral group of $2n$ elements ($n$ odd). So the order in which the product is written actually matters. See also the Note in my answer. $\endgroup$ Jul 26, 2018 at 7:12
  • $\begingroup$ Absolutely, I agree with you @NickyHekster , but doesn't the last part of my post also answer the case when the group is not abelian? Once we have the normality of $H$, the same proof works, right? $\endgroup$
    – Suzet
    Jul 26, 2018 at 13:45
  • $\begingroup$ Actually, I believe it is essentially the same as the proof you gave back in September (which is written in a far more elegant way). But because the OP asked for a "simpler proof", I thought that he would like to avoid talking about Sylow subgroups maybe. That is why I tried to write (rewrite) down something using only elementary notions and manipulations. $\endgroup$
    – Suzet
    Jul 26, 2018 at 14:54
  • $\begingroup$ I could have avoided mentioning the Sylow $2$-subgroup, that is true, since Cauchy"s Theorem provides us with an element $t$ of order $2$. For my Note is it essential. $\endgroup$ Jul 26, 2018 at 16:24

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