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.

  • $\begingroup$ Hint: The order of $G$ is $245=5\cdot 49=5\cdot 7^2$. What does this tell you? $\endgroup$ – Fakemistake Feb 19 '18 at 12:42
  • $\begingroup$ Can you use Sylow theorems to determine the subgroups of $G$? $\endgroup$ – Fakemistake Feb 19 '18 at 12:54
  • $\begingroup$ 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 $G$. $\endgroup$ – Jusko Feb 19 '18 at 13:06
  • $\begingroup$ @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? $\endgroup$ – Jusko Feb 19 '18 at 13:50
  • $\begingroup$ It looks good to me. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ Is it not enough that $H\leq G$ to claim that $b^2\in H$? $\endgroup$ – Jusko Feb 19 '18 at 13:10
  • $\begingroup$ 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). $\endgroup$ – астон вілла олоф мэллбэрг Feb 19 '18 at 13:12
  • $\begingroup$ I should add once again, that having used a very simplistic approach above has really pleased me. $+1$ for your question. $\endgroup$ – 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 $x \in H$.

| cite | improve this answer | |
  • $\begingroup$ Why is $H$ normal if its index in $G$ is $5$? Also, what does $\overline{x}$ means? $\endgroup$ – Jusko Feb 20 '18 at 2:02
  • 1
    $\begingroup$ 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. $\endgroup$ – Nicky Hekster Feb 20 '18 at 14:27
  • 1
    $\begingroup$ 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. $\endgroup$ – Nicky Hekster Feb 20 '18 at 14:29
  • $\begingroup$ 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. :-) $\endgroup$ – Jusko Feb 20 '18 at 14:48
  • $\begingroup$ We bettermives this to a discsussion room $\endgroup$ – Nicky Hekster Feb 20 '18 at 15:48

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.