Consider a group $G$ which is finite and had even order. If we consider the sum of the order of the cyclic groups created by the elements of $G$, is this sum odd? How to prove this?

  • $\begingroup$ i amnot professional in group theory but i think it is because of order of any element divide order of group thus all elements except unit have even order thus sum will be odd $\endgroup$ – R.N Sep 12 '15 at 11:33
  • $\begingroup$ @RaziehNoori what if an element has order 3, so it could still divide let's say the group order 6...that means the order of the elements don't have to be all even $\endgroup$ – user190080 Sep 12 '15 at 11:36
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    $\begingroup$ I'm not sure what you mean - I don't think your question makes it clear what you are supposed to be adding. In relation to one possible interpretation, the cyclic group of order $4$ has cyclic subgroups of size $1,2,4$ sum $7$. The cyclic group order $6$ has subgroups of size $1,2,3,6$ sum $12$. $\endgroup$ – Mark Bennet Sep 12 '15 at 11:37
  • $\begingroup$ @user190080 oh, silly mistake. sorry $\endgroup$ – R.N Sep 12 '15 at 11:37
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    $\begingroup$ The way I read the question: In the cyclic group of order $6$, additive group of $\frac {\mathbb Z}{6 \mathbb Z}$ say, the group generated by $0$ has order $1$, the groups generated by $1$ and $5$ have order $6$, the groups generated by $2$ and $4$ have order $3$, the group generated by $3$ has order $2$...so the sum in question is: $1+2^*6+2^*3+2=21$. $\endgroup$ – lulu Sep 12 '15 at 11:44

The way I read the question, you ask if the sum of the orders of all the elements in your group is always odd. This is true regardless of the parity of the order of the group.

Note that this is true if the group is cyclic (of order $n$). To see this, note that your sum is $$\sum_{d|n} d\varphi(d)$$. Look at each term in that sum. If $d$ is even the term is even, but if $d>1$ is odd then $\varphi(d)$ is even so the only odd term in the sum is $1^*\varphi(1)=1$.

For a general group, with elements $\{e=g_1,g_2, ..., g_n\}$ First look at the group generated by $g_2$. Computing your sum for the elements in that group you get an odd number by our first result. But after that you are summing over the elements in cyclic groups without adding in the identity element (as you already included that in your first sum), hence you always get an even number.

  • $\begingroup$ Can you please explain the notation you have used in the summation. I am used to the one given in the book by Herstein. $\endgroup$ – Almond Pie Sep 12 '15 at 14:07
  • $\begingroup$ sure. the sum is over all the divisors of $n$. $\varphi(n)$ means the euler phi function. Thus, for $n=6$ the divisors are $d=1,2,3,6$ and the sum is $$1^*\varphi(1)+2^*\varphi(2)+3^*\varphi(3)+6^*\varphi(6)=1+2+6+12=21$$ $\endgroup$ – lulu Sep 12 '15 at 14:38
  • $\begingroup$ Wow! Thanks. That helped a lot! $\endgroup$ – Almond Pie Sep 12 '15 at 14:55

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