## Algebra Solutions: Hungerford I.3

1) Let a,b be elements of a group G. Show that $|a| = |a^{-1}|$, $|ab| = |ba|$, and $|a| = |cac^{-1}| \forall c \in G$
Proof-
We show the first equality by showing $\langle a \rangle = \langle a^{-1} \rangle$ and we show this by subset inclusion. Let $g \in \langle a \rangle$. This says that for some $i \in \mathbb{Z}, g = a^i =(a^{-1})^{-i} \in \langle a^{-1} \rangle$. The other direction is similar so that both sets contain each other and thus are the same set, so by definition the order of their generators is the same.
For the second equality, suppose $|ab| = n$. By definition this gives that $e = (ab)^n = ababab...ab = a(ba)^{n-1}b \Rightarrow (ba)^{n-1} = a^{-1}b^{-1} = (ba)^{-1}$ Thus:
$(ba)^n(ba)^{-1} = (ba)^{-1} \Rightarrow (ba)^n=e$. This conclusion tells us that the order of the element ba can be no larger than n, when we are given the fact that the order of ab is n. Thus $|ba| \leq |ab|$ Repeating this argument for $|ba|=m$ we get $|ab| \leq |ba|$ so that $|ab| = |ba|$. The case when either side is infinite automatically gives the other side to be infinite too.
For the last equality, suppose $|a| = n$. Then consider:
$(cac^{-1})^n =cac^{-1}cac^{-1}...cac^{-1} = ca^nc^{-1} = cc^{-1} = e$ Again this tells us the that $|cac^{-1}| \leq n$. Suppose $|cac^{-1}| =m < n$. Then $(cac^{-1})^m =ca^mc^{-1}$ which gives
$e =ca^mc^{-1}$
$c = ca^m$
$e = a^m$ which is a contradiction since the order of a is n. Thus n = m. If either side is infinite then again the other side must be also.

2) Let G be an abelian group with elements a,b having respective orders m,n. Show that G contains an element with order lcm(m,n)
Proof-
First we assume that gcd(m,n)=1. Then we have lcm(m,n) = mn. We have that:
$(ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=ee=e$ so that $|ab| \leq mn$. Suppose $|ab| =k$ Then we have:
$e = e^m= ((ab)^k)^m=(ab)^{km}=a^{km}b^{km}=(a^m)^kb^{km}= b^{km}$. Then since $b^{km} =e$ and the order of b is m, we have that $n | km \Rightarrow n | k$ since gcd(m,n) =1. Similarly $m | k$ Then since gcd(m,n)=1 we have $mn | k$, but $k \leq mn$ so that $k = mn$. Thus the element ab has order lcm(m,n) when gcd(m,n) = 1.
Now suppose that $gcd(m,n) = q > 1$. Then by definition $lcm(m,n) = \frac{mn}{q}$. Note that:
$(a^q)^{\frac{m}{q}} = a^m = e$ and if $(a^q)^r = e$ for some $r < \frac{m}{q}$ then $a^{qr} = e$ where qr < m which contradicts the order of a being m. Thus:
$|a^q| = \frac{m}{q}, |b| = n$ and $gcd(\frac{m}{q},n)=1$. Therefore $|a^qb| = \frac{mn}{q}$ Thus the element $a^qb$ has order lcm(m,n)

3) Let G be an abelian group of order pq with gcd(p,q) = 1. Assume there exists $a,b \in G$ such that $|a| = p$, $|b| =q$ and show that G is cyclic.
Proof-
From number 2 we know that G must contain an element g, with order pq. Because pq is also the order of the group it must be true that g generates G. Thus G is cyclic.

4) If $f:G \rightarrow H$ is a group homomorphism, $a \in G$, and $f(a)$ has finite order in H, then $|a|$ is infinite or $|f(a)|$ divides $|a|$.
Proof-
Let $|f(a)| = m$. Suppose that the order of a is finite, $|a| = n$. Then we have:
$a^n = e \Rightarrow e_H = f(e) = f(a^n) = (f(a))^n$. Then n must divide the order of f(a) so that n divides m.

6) If G is a cyclic group of order n and $k | n$ then G has exactly one subgroup of order k.
Proof-
The case when k is 1 is trivial so assume $k \neq 1$. Let $G = \langle a \rangle$ be cyclic. We can write $n = \frac{n}{k} \cdot k$. This by definition means that $\frac{n}{k} \mid n$. Because $\frac{n}{k} \mid n$ we get that $|a^{\frac{n}{k}}| = \frac{n}{\frac{n}{k}} = k$. Thus $\langle a^{\frac{n}{k}} \rangle$ is a subgroup of G with order k. Now show it is unique. Suppose H is a subgroup of G also with order k. Since k is not 1, H is nontrivial. Being a subgroup of a cyclic group, $H = \langle a^m \rangle$ where m is the least positive integer such that $a^m \in H$ From the division algorithm we have $n=mq+r$ for some $q,r \in \mathbb{Z}$ with $0 \leq r < m$ This gives:
$e = a^n = a^{mq+r} = a^{mq}a^r \Rightarrow a^r = (a^{mq})^{-1} \in \langle a^m \rangle$. However m is the least positive integer such that $a^m \in H$, and we have now that $a^r \in H$ also, with $r which means that r must be 0. Thus $n =mq$ so that $m \mid n$ which gives that gcd(m,n) = m. Then since $gcd(m,n) \mid n$ we have that $|a^{gcd(m,n)}| = \frac{n}{gcd(m,n)} = \frac{n}{m}$ for the same reason as before. Then:
$|H| = k = | \langle a^m \rangle |= |a^m| = |a^{gcd(m,n)}| = \frac{n}{m}$, which gives that $m = \frac{n}{k}$ so that $H = \langle a^{\frac{n}{k}} \rangle$, so the subgroup is unique.

8) A group that has only a finite number of subgroups must be finite.
Proof-
Let a group G have a finite number of subgroups but suppose G is infinite. Let $g_i \in G$ not the identity. Consider $\langle g_i \rangle$. If it is infinite then it must be cyclic and isomorphic to the integers under addition, which has infinitely many subgroups. Thus $\langle g_i \rangle$ cannot be infinite and in fact every element of G must have finite order. Next, let $g_1 \in G$ and again form $\langle g_1 \rangle$. Since G is infinite, but $\langle g_1 \rangle$ is finite we have that $G\backslash \langle g_1 \rangle$ must be infinite as well. Choose another $g_2 \in G\backslash \langle g_1 \rangle$. Form $\langle g_2 \rangle$. Note that $\langle g_1 \rangle \neq \langle g_2 \rangle$. Again we can pick $g_3 \in G\backslash (\langle g_1 \rangle \cup\langle g_2 \rangle )$ and form $\langle g_3 \rangle$. Continuing in this manner we obtain an infinite sequence of subgroups of G which are not all equal. A contradiction to the finite number of subgroups. Thus G is not finite.

9) If G is an abelian group, then the set T of all elements with finite order is a subgroup of G.
Proof-
Since the order of the identity element is 1, $e \in T$ so that T is nonempty. Let $a,b \in T$ have orders m, n respectively. Let lcm(m,n) = q. Then $q = mx = ny$ for some $x,y \in \mathbb{Z}$. Note that $(ab^{-1})^q = a^qb^{-q} = a^{mx}b^{-ny} = e$. This gives that the order of $ab^{-1}$ must divide q, which means the order of $ab^{-1}$ is finite so that $ab^{-1} \in T$ so by the subgroup test, T is a subgroup of G.

10) An infinite group is cyclic iff it is isomorphic to each of its proper subgroups.
Proof-
$(\Rightarrow )$ Assume that G is cyclic so that $G = \langle a \rangle$ for some a in G. Let H be a proper subgroup of G, then we can say that $H = \langle a^m \rangle$, where m is the least positive integer such that $a^m \in H$. Define $\phi : G \rightarrow H$ by $\phi (g) = g^m$. If $g=h \in G$ then $g^m = h^m \in ImG$ so that $\phi$ is well defined.
Let $g \in Ker\phi$. Then $e = \phi (g) = g^m$. Since G is cyclic write $g = a^k$ then since G is infinite, a cannot have finite order so we get:
$e = g^m = a^{km} \Rightarrow km = 0 \Rightarrow k = 0$ since m was positive. Thus $g = e$ so that $\phi$ is injective.
Now for any $h\in H$ we can write $h = a^{mk}$ for some integer k. Note that $a^k \in G$ and this element also maps to h, so $\phi$ is onto.
Lastly, let $g,h \in G$. Then we can write $g = a^k$ and $h = a^n$ so that:
$\phi (gh) = \phi (a^ka^n) = \phi (a^{k+n}) = (a^{k+n})^m = a^{mk+mn} = a^{mk}a^{mn} = \phi (a^k) \phi (a^n) = \phi (g) \phi (h)$ so we have an isomorphism.
$(\Leftarrow )$ Assume that $G \cong H$ for every proper subgroup H of G. Since G is infinite, let a be an element which is not the identity. Then $G \cong \langle a \rangle$, so let $f:\langle a \rangle \rightarrow G$ be an isomorphism. Want to show that $G = \langle f(a) \rangle$ by subset inclusion to show G is cyclic. Let $g\in G$. Since f is onto, there is some $a^k \in \langle a \rangle$ such that $f(a^k) = (f(a))^k =g$ Thus $g\in \langle f(a) \rangle$. Now let $h \in \langle f(a) \rangle$ then $h = (f(a))^k = f(a^k) \in G$ for some integer k. This shows that h is in G. Thus the two sets are equal and so G is cyclic.