# $|G| + \frac{|G|}{\left|\langle a\rangle\right|} + \frac{|G|}{\left|\langle b\rangle\right|} + \frac{|G|}{\left|\langle ab\rangle\right|}$

Show that for every finite group $G$ and for every elements $a, b \in G$ the following expression $$|G| + \frac{|G|}{\left|\langle a\rangle\right|} + \frac{|G|}{\left|\langle b\rangle\right|} + \frac{|G|}{\left|\langle ab\rangle\right|}$$ is even.

• Note ${\rm lcm}(a,b,c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ can easily fail to be even for $a,b,c\in\Bbb N$ so group theory is necessary.
– anon
Commented Nov 21, 2014 at 23:50
• @whacka: will you please explain it more, preferably as an answer? Commented Nov 21, 2014 at 23:59
• @Krish my comment is not an answer, it simply says that an answer will need to use more than just pure arithmetic, in case anybody was trying to go down that route.
– anon
Commented Nov 22, 2014 at 0:21
• Very weird. What about cyclic groups? If $n,a,b$ are positive integers such that $a$ and $b$ divide $n$, why is $n + a + b + \mathrm{gcd}(a+b,n)$ even? Commented Nov 22, 2014 at 2:46
• I wonder the source of the quesion, can you please write it ? Commented Nov 22, 2014 at 11:43

I think I got it. Check it please.

1. If $|G|$ is odd. Obvious.

2. $|G|$ is even. Then let's assume embedding $f:G \rightarrow S_{|G|}$. According to Cayley theorem $g$ maps to the product of $\frac{|G|}{\left|\langle g\rangle\right|}$ independent cycles of length $\left|\langle g\rangle\right|$. In particular $g$ maps to the odd permutation iff $|G|$ is even and $\frac{|G|}{\left|\langle g\rangle\right|}$ is odd.

3. There are either $2$ or $0$ odd permutations among $f(a), f(b), f(ab)$, i.e. there are zero or two odd summands in this sum ($|G|$ is even).

• That looks right to me. Commented Nov 22, 2014 at 13:43
• Very nice. Where did this problem come from?
– anon
Commented Dec 31, 2014 at 4:24

I think the following works.

If the order of $G$ is odd, we are done, so suppose that $G$ has even order.

If $[G:\langle x\rangle]$ is even, for $x\in\{a,b,ab\}$, we are done, so suppose that $[G:\langle x\rangle]$ is odd for some $x\in\{a,b,ab\}$. The $\langle x\rangle$ contains a Sylow $2$-subgroup $P$ of $G$. Since $P$ is cyclic, it follows that $G$ has a normal Hall $2^\prime$-subgroup $Q$. Then $G = PQ$ and $P\cap Q=1$.

Now consider the quotient group $\bar{G} = G/Q$, which is a non-trivial cyclic $2$-group isomorphic to $P$. We have, by the correspondence theorem, $$\left|G\right|+[G:\langle a\rangle]+[G:\langle b\rangle]+[G:\langle ab\rangle] \\ = \left|Q\right|\left|\bar{G}\right| + [\bar{G}:\langle\bar{a}\rangle]+[\bar{G}:\langle\bar{b}\rangle]+[\bar{G}:\langle\overline{ab}\rangle].$$ Now all the terms on the right hand side are even, unless $\bar{G}$ is generated by one of $\bar{a}$, $\bar{b}$ or $\overline{ab}$. But, it is easy to see that, in each case, the resulting sum is even nonetheless.

For example, suppose that $\bar{G} = \langle\bar{a}\rangle$, so that $[\bar{G}:\langle\bar{a}\rangle]=1$. Then $\bar{b} = \bar{a}^r$, for some $r$, and so $\overline{ab} = \bar{a}^{r+1}$ and we get $$\left|G\right| + 1 + [\bar{G}:\langle \bar{a}^r\rangle] + [\bar{G}:\langle\bar{a}^{r+1}\rangle].$$ Since $\bar{a}$ has order a power of $2$, and one of $r$ and $r+1$ is odd, we must have $\langle\bar{a}^r\rangle=\langle\bar{a}\rangle$ or $\langle\bar{a}^{r+1}\rangle=\langle\bar{a}\rangle$, so one of the indices $[\bar{G}:\langle\bar{a}^r\rangle]$ and $[\bar{G}:\langle\bar{a}^{r+1}\rangle]$ is also equal to $1$, while the other is even, so the result is even.

• I missed some point. How can you say that when $P$ is cyclic then it has a normal complement ? Commented Nov 22, 2014 at 11:27
• Hi @mesel. If a Sylow $p$-subgroup $P$ of a finite group $G$ is cyclic, where $p$ is the least prime divisor of the order of $G$, then it has a normal $p$-complement. This is an application of the transfer homomorphism into $P$. See the sections on the transfer in the books by Robinson or Rotman. Commented Nov 23, 2014 at 23:56
• Hi @james. Interesting fact. Thank you. Commented Nov 24, 2014 at 0:45

I have an idea for soluble groups. But I can not find a way for non soluble groups.

If $|G|$ is odd, then it is done. So we assume that $|G|$ is even. Let $|G|=2^r f$. We need only to consider that $G=<a,b>$.

Case 1. $G$ is abelian. Suppose $|a|=2^r m, |b|=2^4 n, |ab|=2^r l$. Then, by the uniqueness of the Sylow 2-subgroup of abelian groups, we get $<x>=<a^m>=<b^n>$, which is the Sylow s-subgroup of $G$. We can assume that $x=a^m=b^{n'}$, where $m, n'$ are all odd. Now $x \in <ab>$. So $x=(ab)^t$. We get $t$ is odd. but now $a^t=b^{n'-t}$ and $b^t=a^{m-t}$. Since $n'-t$ and $m-t$ are both even now, we see that the 2-part of $|a^t|$ and $|b^t|$ are $\le 2^{r-1}$, which means that the 2-pare of $|x|=|a^t b^t|$ is $\le 2^{r-1}$, a contradiction. So $|G:<a>|, |G:<b>|, |G:<c>|$ can not be all odd. We need to consider the case that $|G:<a>$ and $|G:<b>|$ are even, but $|G:<ab>|$ is odd. But this does not occur because the 2-part of $|ab|$ is $\le$ that of $|a|$ and $|b|$.

Case 2. If $G'<G$, we can consider $\bar{G}=G/G'$ to get the required result.

Hence this is true for soluble groups.