Let $G$ be a group of order 245. If $G$ has a subgroup $H$ of order $49$, prove that every element of order $7$ is in $H$.
What I know: The order of a subgroup divides the order of the group if the group is finite, which is the case.
I can show that there is only one subgroup of $G$ that is of order $49$, $H$ is unique. What else do I need to show in order to prove that every element of $G$ of order $7$ is in $H$?
Thank you in advance for your time and help.
On
Other approach: if you assume that there exists a subgroup $H$ of $G$, with $|H|=49$, then $|G:H|=5$ the smallest prime dividing $|G|$, and hence $H \lhd G$. If $x \in G$ with o$(x)=7$, then applying Lagrange's Theorem in $G/H$ we have $\bar{x}^7=\bar{x}^5 \cdot \bar{x}^2=\bar{x}^2=\bar{1}$. Since o$(\bar{x})$ divides also $5$, we get $\bar{x}=\bar{1}$, that is $x \in H$.
Let $b$ be an element of order $7$ in $G$. Suppose that $b \notin H$. Note that $b^2 \notin H$, for if it were in $H$, then so would $(b^2)^4 = b$.
Similarly, note that $b^k \notin H$ for all $1 \leq k \leq 6$, since given any such $k$, we have a unique $1 \leq k' \leq 6$ such that $kk'$ leaves a remainder of $1$ upon division by $7$, and then $b^k \in H \implies (b^k)^{k'} = b \in H$, a contradiction.
Now, note that therefore, $b^kH, 1 \leq k \leq 6$, are distinct cosets of $H$ in $G$, since if two of them were the same, this would show that $b^{k-l} \in H$ for some $1 \leq k,l \leq 6$, forcing $k=l$.
But then, there are only $5$ cosets of $H$ in $G$, since the number of cosets is just $\frac{|G|}{|H|} = \frac{245}{49} = 5$. This gives a contradiction to the previous statement, forcing $b \in H$.
I don't think I used anything about the size of $G$ and $H$, other than the fact that their ratio is $5$ here. So this may work in a more general framework.