- Thread starter
- Banned
- #1

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

- Admin
- #2

- Jan 26, 2012

- 4,202

Not true, if I'm understanding correctly. Take $\mathbb{Z}_{20}$, and the subgroups $\langle 5\rangle$ and $\langle 10\rangle$. Then $y=5\not\in\langle 10\rangle$, but $\langle 5\rangle\cap\langle 10\rangle =\{0,10\}$.

I think you have to be talking about these things in the context of some larger group, and the order of that group matters. If $|\langle x\rangle|=p$ and $|\langle y\rangle|=q$, where $p\not=q$ are both primes (might even work if they are only relatively prime), and if the order of the larger group is equal to $pq$, then it would be true.

I think you have to be talking about these things in the context of some larger group, and the order of that group matters. If $|\langle x\rangle|=p$ and $|\langle y\rangle|=q$, where $p\not=q$ are both primes (might even work if they are only relatively prime), and if the order of the larger group is equal to $pq$, then it would be true.

Last edited:

- Thread starter
- Banned
- #3

Thanks

- Moderator
- #4

- Feb 7, 2012

- 2,799

Suppose that every non-identity element of G has order p (prime). Then $\langle x\rangle = \{x^k:1\leqslant k\leqslant p\}$, and similarly for $y$. Suppose that $\langle x\rangle$ and $\langle y\rangle$ have a non-identity element in common, say $y^r=x^s.$ Let $t$ be the inverse of $r$ in $\mathbb{Z}_p^\ast$. Then $x^{st} = y^{rt} = y$, so that $y\in \langle x\rangle.$ Contrapositively, if $y\notin \langle x\rangle$ then $\langle x\rangle$ and $\langle y\rangle$ have no non-identity element in common.Yes it must work if p and q are co-prime because of the result that the order of an element divides the order of a group. If it doesn't work in the case p=q=prime, then I am confused about something in my textbook. We have a group G of order prime squared, non cyclic and it can be proved that the centre is non-trivial. Then we may choose x in Z(G) not the identity and y not in <x>. The book then states that <x> and <y> have only the identity in common. But clearly every non-identity element has order p.

- Thread starter
- Banned
- #5

- Feb 15, 2012

- 1,967

the key here is that the order of a subgroup is severely restricted:

we can only have non-trivial proper subgroups of order p.

now any non-identity element of a group of order p (which is necessarily cyclic) generates the entire group.

so if x is an element of order p, and y is in <x>, either:

y = e, or:

<x> = <y>.

if y ≠ e is NOT in <x>, then <x> and <y> must be distinct subgroups of order p, thus

<x> ∩ <y> can only have order p, or order 1.

if <x> ∩ <y> has order p, then <x> = <y> which means y is in <x>, a contradiction.

so <x> ∩ <y> must have order 1, so the intersection is trivial.

as an aside, i remark that:

suppose we have x in Z(G), and choose y not in <x>,

since x is central, we have that x commutes with all of <y>, and thus all of <x> commutes with all of <y>.

since <x> ∩ <y> = {e}, <x><y> = {x^{j}y^{k}: 0 ≤ j,k ≤ p-1} is all of G, and one can show directly that this group is therefore abelian:

(x^{j}y^{k})(x^{j'}y^{k'}) = x^{j}(y^{k}x^{j'})y^{k'} = x^{j}(x^{j'}y^{k})y^{k'}(since all of <x> commutes with all of <y>)

= x^{j+j'}y^{k+k'} = x^{j'+j}y^{k'+k} = x^{j'}(x^{j}y^{k'})y^{k} = x^{j'}(y^{k'}x^{j})y^{k} = (x^{j'}y^{k'})(x^{j}y^{k})

we can only have non-trivial proper subgroups of order p.

now any non-identity element of a group of order p (which is necessarily cyclic) generates the entire group.

so if x is an element of order p, and y is in <x>, either:

y = e, or:

<x> = <y>.

if y ≠ e is NOT in <x>, then <x> and <y> must be distinct subgroups of order p, thus

<x> ∩ <y> can only have order p, or order 1.

if <x> ∩ <y> has order p, then <x> = <y> which means y is in <x>, a contradiction.

so <x> ∩ <y> must have order 1, so the intersection is trivial.

as an aside, i remark that:

suppose we have x in Z(G), and choose y not in <x>,

since x is central, we have that x commutes with all of <y>, and thus all of <x> commutes with all of <y>.

since <x> ∩ <y> = {e}, <x><y> = {x

(x

= x

Last edited: