2
$\begingroup$

I am not sure about this. I understand it when it is cyclic. But if not stated so, I cannot reason as to why.($G$ here I assume finite)

$G$ is a group with some subgroup $H$. Then, if $|H|=n$ then there is an element $g \in G$ with order $n$.

Is this true? Then why? I ask this because depending on the sources, when I look at Lagrange's theorem they use it sometimes to say that "there is an element of order $n$ when $|H|=n$ exists in $G$" or indicates so.

I mean, if $H$ is cyclic and has $n$ elements, then it simply means the generator $g$ will have to be multiplied $n$ times to reach around all elements, so $g^n=e$ is understandable. But otherwise...I'm not so sure. Is there a theorem about this?

Improve this question
$\endgroup$
  • $\begingroup$ What's the relationship between $G$ and $H$? Is one a subgroup of the other? $\endgroup$ – user307169 May 12 '16 at 11:08
  • $\begingroup$ $A_4$ is a subgroup of $S_4$ of order $12$ but there is no element of order $12$ in $A_4$ $\endgroup$ – Kushal Bhuyan May 12 '16 at 11:12
  • $\begingroup$ ugh, sorry, I missed out the "sub" $\endgroup$ – Kydo May 12 '16 at 11:13
2
$\begingroup$

No. $V_4$ is a group of order $4$ with no element of order $4$.

Improve this answer
$\endgroup$
  • $\begingroup$ Alright, but then can you explain this line from a proof "if $|G|=p^n$ then for any $g \in G$ its order divides $p^n$ by Lagrange" which is given in a proof of a corollary in here torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf I don't understand how the order of the group necessarily correlates to the order of an element of the group $\endgroup$ – Kydo May 12 '16 at 11:18
  • 1
    $\begingroup$ Lagrange theorem states that if $G$ is a group and $H$ is any subgroup of $G$, then the order of $H$ divides the order of $G$. So given any element $g\in G$, take $n$ the order of the element and consider $\{g, g^2,\dots, g^n=e\}$, where $e$ is the neutral element of $G$. This is obviously a subgroup of $G$ since it is closed for the operation in $G$. Hence we have a subgroup of $g$ of order the order of $g$, so the order of $g$ divides the order of the group. $\endgroup$ – David Méndez May 12 '16 at 11:24
1
$\begingroup$

The order of a subgroup is simply the number of elements in the subgroup. For example

$$A_n \leq S_n$$

with $A_n$ a subgroup of order $n!/2$ in the group $S_n$ of order $n!$.

Lagrange's Theorem states that the order of a subgroup divides the order of the group, as you can see this is true in the above example.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.