If $G$ is a cyclic group of order $n$ and $k|n$, then $G$ has a subgroup of order $k$, it's easy to prove, but what I'm having troubles to show is there is only one subgroup of order $k$.
I need help here
Thanks a lot
[Math] In a finite cyclic group there is only one subgroup of order $k$.
abstract-algebragroup-theory
Related Solutions
At this level, there are many things you have to show. Let's do it in small steps.
- First show that $ab$ does not have order $1$, i.e. that $ab \neq e$. To do this, assume that $ab = e$ and derive a contradiction to the assumption that $a \neq b$. Remember that $b^2 = e$.
- Now you want to show that $(ab)^2 = e$. Write out $(ab)^2 = abab$ and remember that the group is commutative, so in particular $a$ and $b$ commute.
- Show that $H$ is actually a subgroup. It contains the identity element of $G$, so that's good. Write out a multiplication table to prove that $H$ is closed under the group operation. This will also show that $H$ contains inverses for all of its elements.
- Show that $H$ has order $4$. We know that neither $a$ nor $b$ is equal to $e$ because $e$ has order $1$, and we assumed that $a\neq b$, so $\{e, a, b\}$ has three elements. You showed in (1) that $ab \neq e$, and now show that $ab = a$ and $ab = b$ both lead to contradictions, and hence that $\{e, a, b, ab\}$ has $4$ elements.
- Check that each element of $H$ generates a subgroup of $H$ of order less than $4$. Remember what this means: an element $x$ of a group generates a cyclic subgroup consisting of all the powers of $x$: \[ \ldots, x^{-2}, x^{-1}, e, x, x^2, \ldots \] If $x$ has finite order $n$, then there will only be finitely many elements in this list, and in fact $\{e, x, \ldots, x^{n - 1}\}$ is the complete list of distinct elements of the subgroup generated by $x$. If you haven't proved this, then you should! Thus $a$ generates a subgroup of order $2$, and so on.
It's also fun to prove that the cyclic group of order $4$ and this group $H$ (which is known as the Klein four-group) are the only groups of order $4$ up to isomorphism.
What might make one think that there may be more than one subgroup of order 5?
Experience: it doesn't take long, as you examine small groups, before you run into groups that have multiple subgroups of the same size: $S_3$, the nonabelian group of order $6$ that corresponds to the bijections of $\{1,2,3\}$ with itself (under composition) has three different subgroups of order $2$. The Klein $4$-group, which is $C_2\times C_2$ (direct product of two cyclic groups of order $2$) has three subgroups of order $2$. The dihedral group with $8$ elements, $D_8$, has five subgroups of order $2$. $C_3\times C_3$, the direct product of two cyclic groups of order $3$, has four subgroups of order $3$; etc. In fact, it's pretty rare for a subgroup $H$ of a group $G$ to be the only subgroup of size $|H|$, except when $|H|=1$ or $|H|=|G|$. It does happen, of course: it happens with cyclic groups. In the alternating group $A_4$ (which has 12 elements), there is a unique subgroup of order $4$. In $S_3$, there is a unique subgroup of order $3$. And so on.
What are the criteria for there being only one [subgroup of a given order] rather than more than one?
Well, of course there's the obvious: $H$ is the only subgroup of $G$ of order $|H|$ if and only if for every subgroup $K$ of $G$, either $K=H$ or $|H|\neq |K|$.
There are some sufficient conditions: if $G$ is finite and cyclic, then $G$ contains a unique subgroup of order $d$ for each $d$ that divides $|G|$. (You may not be aware of this result yet; you'll get to it pretty soon).
But in general it is pretty hard to give a criteria. Essentially, you either just show that the given subgroup is the only one with that order (usually in some relatively ad hoc manner), or you exhibit two distinct subgroups of the same order.
Now, on to the proof. First we show that $G$ must have at least one subgroup of order $5$, no matter what.
Let $g\in G$, $g\neq e$; since $|G|=25$, we can certainly pick an element that is not the identity. By Lagrange's Theorem, the order of $g$ must divide $25=5^2$, so the order of $g$ is either $1$, $5$, or $25$. It cannot be $1$, because the only element of order $1$ is the identity, and we explicitly excluded the identity from our choice of $g$.
If the order of $g$ is $5$, then let $H=\langle g\rangle$. The subgroup generated by $g$ contains exactly as many elements as the order of $g$, so $|H|=\mathrm{order}(g)=5$; and of course, "the subgroup generated by $g$" is a subgroup of $G$. So if the order of $g$ is $5$, then we are done.
The only other possibility is that the order of $g$ is $25$. That means that $g^n=e$ if and only if $25|n$. I claim that the order of $g^5$ is $5$: first, notice that $(g^5)^5= g^{25} = 1$, so the order of $g^5$ divides $5$ (remember that if $x^n=e$, then the order of $x$ divides $n$). So the order of $g^5$ is either $1$ or $5$; it cannot be $1$, because that would mean that $e = (g^5)^1 = g^5$, which would imply, in turn, that the order of $g$ divides $5$. But we are assuming that the order of $g$ is $25$, so $g^5\neq e$, so the order of $g^5$ is not $1$, so the order of $g^5$ must be $5$. Now taking $H=\langle g^5\rangle$ gives a subgroup of order $5$, just as above.
So if $|G|=25$, then there is a subgroup $H$ of $G$ of order $5$.
(In fact, the above argument can be made, exactly the same way, if $|G|=p^2$ for any prime $p$: any such group must have a subgroup of order $p$).
Now on to the second part: we want to prove that our $G$ has a unique subgroup of order $5$ if and only if $G$ is cyclic. This means that we have two prove two implications: if $G$ is cyclic then it has a unique subgroup of order $5$; and if $G$ has a unique subgroup of order $5$ then $G$ is cyclic.
First, we prove that if $G$ is cyclic then it has a unique subgroup of order $5$: since $G$ is cyclic, there is an $x\in G$ such that $G=\langle x\rangle = \{e,x,x^2,\ldots,x^{24}\}$. If $K$ is any subgroup of $G$, other than the trivial subgroup, then there is a smallest positive integer $k$, $1\leq k \lt 25$, such that $x^k\in K$. I claim that $k$ divides $25$ and that $K=\{e,x^k, x^{2k},\ldots,x^{(q-1)k}\}$, where $25 = qk$.
Indeed, using the division algorithm, we can divide $25$ by $k$ with remainder, $25 = qk + r$, $0\leq r\lt k$. Then $$e = x^{25} = x^{qk+r} = x^{qk}x^r = (x^k)^qx^r.$$ Therefore, $x^r = \left((x^k)^q\right)^{-1}$. Note that $x^k\in K$ and $K$ is a subgroup, so $(x^k)^q\in K$, and hence $x^r\left((x^k)^q\right)^{-1}\in K$. But $k$ is the smallest positive integer such that $x^k\in K$, and $0\leq r\lt k$. Since $x^r\in K$, we conclude that the only way this can happen is if $r=0$; and therefore, $25 = qk$, so $k$ divides $25$. And then $k$, $2k,\ldots, (q-1)k$ are all positive and strictly smaller then $25$, so $x^k$, $x^{2k},\ldots,x^{(q-1)k}$ are all distinct. Since $x^{qk} = x^{25}=e$, we get that $K=\{e,x^k,x^{2k},\ldots,x^{(q-1)k}\}$. In particular, $K$ has order $25/k$.
Now, let $H$ be a subgroup of $G$ of order $5$. If the smallest positive $k$ such that $x^k\in H$, then $H$ has order $25/k = 5$. That means that $k=5$, so $H=\{e,x^5,x^{10},x^{15},x^{20}\}$. Note that we concluded this merely from the assumption that $H$ has order $5$; that means that every subgroup of $G$ of order $5$ is equal to $\{e,x^5,x^{10},x^{15},x^{20}\}$, so $G$ has a unique subgroup of order $5$. This proves one of the two implications.
To finish off, we need to show that if $G$ has a unique subgroup of order $5$, then $G$ is cyclic. Let $x\in G$, $x\neq e$. As before, we know that the order of $x$ is either $5$ or $25$; if it is $25$, we are done, because then $\langle x\rangle = G$ and $G$ is cyclic. So assume that $x$ has order $5$. Then $H=\langle x\rangle$ is the unique subgroup of $G$ of order $5$, by assumption. Now, $G-H$ is not empty (there's 20 things in it); let $y\in G-H$. Again, the order of $y$ must be either $1$, $5$, or $25$: it cannot be $1$, because the only element of order $1$ is the identity, and the identity is in $H$, whereas $y\in G-H$. The order cannot be $5$ either: if the order were $5$, then $K=\langle y\rangle$ would be a subgroup of $G$ order $5$; by our assumption $G$ has a unique subgroup of order $5$, so that would mean that $K=H$, since both have order $5$; but then $y\in \langle y\rangle = K = H$, again contradicting our choice of $y$ as an element of $G-H$.
But that only leaves one possibility: that the order of $y$ is $25$. But if the order of $y$ is $25$, then $\langle y\rangle = G$, so $G$ is cyclic, as claimed.
(Again, the argument above works for any group of order $p^2$, where $p$ is a prime; the only thing we used about $25$ is that the only divisors are $1$, $5$, and $25$).
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Best Answer
Consider the epimorphism $f\colon\mathbb Z\to G$ given by sending $1$ to a chosen generator of $G$. Then $\ker f=n\mathbb Z$. Show that for $H<G$ we have $f^{-1}(H)=m\mathbb Z$ (as that's how subgroups of $\mathbb Z$ look like) for some $0<m|n$ and that necessarily $m=[G:H]$.