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

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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.