# Suppose $G$ is a group with exactly $8$ elements of order $10$. How many cyclic subgroups of order $10$ does $G$ have?

In a cyclic subgroup of order $$10$$, there are $$\phi(10)=4$$ elements of order $$10$$.
Since there are exactly $$8$$ elements of order $$10$$, we can choose $$4$$ elements out of the $$8$$ elements of order $$10$$ in $$^8C_4=70$$ ways and for each way we have a cyclic subgroup of order $$10$$ ($$4$$ elements will of order 4 and the rest 4 will come from group G). Thus, $$G$$ has $$70$$ cyclic subgroups of order $$10$$.

But the answer to this question is $$2$$ since a cyclic subgroup of order $$10$$ can only have $$4$$ elements of order $$10$$ and hence there are $$8/4$$ cyclic subgroups only. But I don't understand why? There are $$^8C_4=70$$ ways to choose $$4$$ elements out of $$8$$ elements. So I think it should be $$70$$.

PS: I know that in a finite group, no. of subgroups should divide the order of the group. However, in my case since order of $$G$$ is not given, I am confused.

• Why do you suppose each of the $\binom{8}{4}$ ways of choosing four of the elements corresponds to a different subgroup? Let one of the eight elements be called $a$. Consider the subgroup generated by $a$... so you have $0, a, a+a, a+a+a, a+a+a+a, \dots, 9\cdot a$. Now... notice that $a, 3a, 7a,$ and $9a$ are all of order $10$. – JMoravitz Jun 1 '20 at 13:08
• For building intuition, consider $\Bbb Z_{10}\times \Bbb Z_{10}$ – JMoravitz Jun 1 '20 at 13:08
• @JMoravitz, I think I am starting to understand it now. If $a\in G$ is of order $10$ then $\{e,a,a^2,...,a^9\}$ is a subgroup of order 10 and it does have exactly $4$ elements of order 10 viz. $a,a^3,a^7,a^9$. Now $4$ are left to be utilized. – Koro Jun 1 '20 at 13:15
• I don't think it is possible for a group to have exactly eight elements of order $10$, which would mean that the question was meaningless. – Derek Holt Jun 1 '20 at 13:35
• @Derek Holt, it's an exercise problem from Gallian's abstract algebra. – Koro Jun 1 '20 at 13:39

I claim that there is no group with exactly $$8$$ elements of order $$10$$, which makes the whole question logically meaningless (except possibly as a lemma in the proof that there is no such group).

Such a group $$G$$ would have two exactly two distinct subgroups $$A$$ and $$B$$ of order $$10$$. If they were not normal in $$G$$, then they would be conjugate and their normalizer would have index $$2$$ and contain both $$A$$ and $$B$$, so we could replace $$G$$ by this subgroup to get a smaller example.

So we can assume that $$G = \langle A,B \rangle = AB$$ with $$A,B \unlhd G$$.

Since $$|{\rm Aut}(A)| = 4$$, the element $$g$$ of order $$5$$ in $$B$$ must centralize $$A$$. Then, if $$g \not\in A$$, then $$\langle g, A \rangle \cong C_5 \times C_{10}$$ has $$6$$ subgroups of order $$15$$, contrary to assumption.

So $$g \in A$$, $$|A \cap B| = 5$$, and $$|G| = |AB| = 20$$.

Now, of the five (isomorphism types of) groups of order $$20$$, $$C_{20}$$, $$D_{20}$$ and the dicyclic group have a unique subgroup of order $$10$$, $$C_2 \times C_{10}$$ has three such subgroups, and the final group, which is a semidirect product $$C_5 \rtimes C_4$$ with faithful action has none.

• His question makes sense you've proven no such object exists. – cactus314 Jun 1 '20 at 20:09
• @cactus314 Yes, but the statement "All groups that have exactly eight elements of order 10 have 263 subgroup of order 10" is true! – Derek Holt Jun 1 '20 at 21:32
• Can I ask a clarification question? When you write $G=\langle A, B\rangle$, is this a different $G$ from the normaliser? I cannot see any reason why the two subgroups would generate the normaliser. (The proof only uses that $G$ is a subgroup of the normaliser, so this isn't an error, it's just unclear to me what you're meaning.) – user1729 Jul 18 at 10:33
• @user1729 I believe that the argument proves that, if $G$ is a minimal counterexample to the claim that no such group exists, then $G = \langle A,B \rangle$ and $A,B \unlhd G$. So I think the answer to your question is that we have proved (under this assumption) that $G = \langle A,B \rangle$ and $A,B \unlhd G$. So yes that means that $G$ is equal to the normalizer. The problem with all this is that this group $G$ doesn't exist anyway, so it is difficult to be definitive about its properties! – Derek Holt Jul 18 at 11:32

As you have correctly pointed out that each cyclic subgroup will have exactly 4 elements of order 10.

claim:- if $${H_k}\cap{H_l}$$ contains an element (say) $$a_i$$ of order 10 then $$H_k=H_l$$

Since $${a_i}\in{H_k}\cap{H_l}$$ and order of $$a_i$$ is 10 then, $$=H_k$$ and $$=H_l$$ Hence $$H_k=H_l$$

Did you got why there will be only two cyclic subgroup of order 10?