# Does “order of a subgroup $n$” mean “there is an element of order $n$ in $G$”?

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?

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

No. $V_4$ is a group of order $4$ with no element of order $4$.
• 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 – Kydo May 12 '16 at 11:18
• 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. – David Méndez May 12 '16 at 11:24
$$A_n \leq S_n$$
with $A_n$ a subgroup of order $n!/2$ in the group $S_n$ of order $n!$.