As per Lagrange's theorem, the order of a subgroup must perfectly divide the order of the group.
Let us take the group $G = (S, \sharp)$, where $S = \{1,3,5,7,9\}$ and
$\sharp$ suggests multiplication mod 10. $G$ is a group, with being
$O(G)=5$
Let us take a subset $H = \{1,3,7,9\}$, then $(H, \sharp)$ is also a group as it is closed, associative, satisfies identity and inverse laws. Thus $H$ is a subgroup of $S$. $O(H) = 4$.
But I don't see Lagrange's theorem holding, as 4 does not divide 5.
Not sure if I'm making any silly mistake somewhere, would appreciate some help on this.
Best Answer
The original $G$ is not a group:
$$5~\sharp~5 = 25 \mod 10 = 5$$
If $5$ had an inverse then by multiplying by $5^{-1}$ on both sides we get:
$$5=(5^{-1}~\sharp~5)~\sharp~5=5^{-1}~\sharp~(5~\sharp~5) = 5^{-1}~\sharp~5 = 1$$
Which is clearly not correct. Hence $5$ doesn't have an inverse in $G$.