# Prove that every element of $G$ of order 7 is in $H$.

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 $H$?

• Hint: The order of $G$ is $245=5\cdot 49=5\cdot 7^2$. What does this tell you? – Fakemistake Feb 19 '18 at 12:42
• Can you use Sylow theorems to determine the subgroups of $G$? – Fakemistake Feb 19 '18 at 12:54
• Is there other way aside from using the Sylow theorem? I was thinking that the orders of the subgroups of $G$ is a divisor of the order of $G$. – Jusko Feb 19 '18 at 13:06
• @Fakemistake By the First Sylow Theorem, every subgroup of $G$ of order $7$ is a normal subgroup of of $H$. Then it follows that $\langle a \rangle \leq H$ so that $\langle a \rangle \subseteq H$. Thus, $a \in H$. Am I correct? – Jusko Feb 19 '18 at 13:50
• It looks good to me. – Fakemistake Feb 19 '18 at 14:44

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.

• Is it not enough that $H\leq G$ to claim that $b^2\in H$? – Jusko Feb 19 '18 at 13:10
• No, I do not think it is enough that $H \leq G$ to claim $b^2 \in H$. But then, I may be wrong, because I do not know Sylow's theorem etc. very well, and it may be possible that with the structure theorem or so such a conclusion can be derived. As far as I know, $H \leq G$ is not enough to claim $b^2 \in H$(If it were, though, then the answer is even more obvious than I've written, since if $b^2 \in H$ then of course $b \in H$ using the multiplication argument). – астон вілла олоф мэллбэрг Feb 19 '18 at 13:12
• I should add once again, that having used a very simplistic approach above has really pleased me. $+1$ for your question. – астон вілла олоф мэллбэрг Feb 19 '18 at 13:18

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 $x \in H$.

• Why is $H$ normal if its index in $G$ is $5$? Also, what does $\overline{x}$ means? – Jusko Feb 20 '18 at 2:02
• In general if in a finite group $G$ for a subgroup $H$ the index $|G:H|=p$, the smallest prime $p$ dividing $|G|$, then $H$ is normal. I can provide you with at least two different proofs. You can also look it up here at StackExchange since it was the subject of many posts. – Nicky Hekster Feb 20 '18 at 14:27
• The bar notation $\bar{x}$ designates that I am considering $x$ as an element in the factor group $G/H$ via the canonical projection, so $\bar{x}=xH$ as coset notation. Left or right coset does not matter, $H$ is normal. – Nicky Hekster Feb 20 '18 at 14:29
• Such a great help. Sorry for this request but can you further help me through your application of the Lagrange's Theorem in the factor group. :-) – Jusko Feb 20 '18 at 14:48
• We bettermives this to a discsussion room – Nicky Hekster Feb 20 '18 at 15:48