Corollaries of Lagrange's Theorm

[PsychonautQQ]

Homework Statement

Not a homework question actually. I'm having trouble understanding some of the corollaries to Lagrange's theorem.

Theorem: Let H b e a subgroup of a finite group G. then |H| divides |G|

Corollary 1: if g is an element of a finite group G, then |g| divides |G|.
proof: the cyclic subgroup |H| = <g> generated by g has |H| = |g|.
question: how do we know such a cyclic subgroup H exists as required by the proof?

Corollary 2: If p is a prime, then every group G of order p is prime.
proof: write H = <g>. Then |H| divides |G| so |H| is 1 or |H| = p = |G|.
question: Again, how do we know that such a cyclic subgroup H exists in the first place?

 

[jbunniii]

Corollary 1: if g is an element of a finite group G, then |g| divides |G|.
proof: the cyclic subgroup |H| = <g> generated by g has |H| = |g|.
question: how do we know such a cyclic subgroup H exists as required by the proof?

What is your definition of <g>?

Corollary 2: If p is a prime, then every group G of order p is prime.

I assume you mean "is cyclic".

proof: write H = <g>. Then |H| divides |G| so |H| is 1 or |H| = p = |G|.
question: Again, how do we know that such a cyclic subgroup H exists in the first place?

Same question as above: what is your definition of <g>?

 

[PsychonautQQ]

<g> is defined a subgroup of G where g is an element of G and all the elements of <g> are {1, g, g^2, g^3, g^(n-1} where |G| = n. I believe.

Yes I meant to say "is cyclic" not "is prime" my bad.

 

[jbunniii]

<g> is defined a subgroup of G where g is an element of G

So far so good.

and all the elements of <g> are {1, g, g^2, g^3, g^(n-1} where |G| = n. I believe.

$|G| = n$ is only true if $G$ is cyclic (and finite), and $g$ is a generator for $G$. More generally, $\langle g\rangle$ might be a proper cyclic subgroup of $G$. However, the key is that it's a subgroup. So what does Lagrange's theorem tell you about the size of $\langle g\rangle$?

 

[PsychonautQQ]

so this corollary only works if G is cyclic thus has a cyclic subgroup? or this corollary only works assuming there is a cyclic subgroup <g> of G? I feel like something is being assumed that the corollary statement didn't outright state.

 

[jbunniii]

so this corollary only works if G is cyclic thus has a cyclic subgroup? or this corollary only works assuming there is a cyclic subgroup <g> of G? I feel like something is being assumed that the corollary statement didn't outright state.

$\langle g \rangle$ is always a cyclic subgroup, even if $G$ is not.

 

[PsychonautQQ]

so every group has cyclic subgroups... I guess that's obvious now that I think about it. thanks a ton man, I'm new to this stuff X_X

 

[jbunniii]

so every group has cyclic subgroups... I guess that's obvious now that I think about it. thanks a ton man, I'm new to this stuff X_X

Yes, that's right. Every element $g$ generates a cyclic subgroup $\langle g \rangle$, but not necessarily distinct subgroups - some elements might generate the same subgroup. (For example, $\langle g \rangle = \langle g^{-1} \rangle$.)

Since $\langle g \rangle$ is a subgroup, Lagrange's theorem constrains its size to be a divisor of $|G|$.

The converse is not true in general: if $d$ is a divisor of $|G|$, there isn't necessarily a subgroup (cyclic or not) of order $d$.

However, pretty soon you will probably learn Cauchy's theorem, which says that if $p$ is any prime divisor of $|G|$, then $G$ has a cyclic subgroup of order $p$.