# Two group theory questions

I have two questions here about this topic that have absolutely baffled me. Assistance in solving these would be appreciated.

Q1:

Let $S := \{(123),(235)\} \subseteq \Sigma_5$. Determine the subgroup $\langle$$S$$\rangle$ $\leq$ $\Sigma_5$.

Q2:

Let $G$ be a finite group with subgroups $A,B \leq G$. Show that $$[G : A\cap B] = [G : A] \centerdot [A : A\cap B] \leq [G : A] \centerdot [G : B].$$

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Q1: note that $4$ is a fixed point, so simply work in $\Sigma_4$. Q2: makes no sense. What is $A/B$ and how is a subgroup of $G$ (or $A$)? – user641 Feb 25 '12 at 16:56
For Q1 could be usefull considering that $\langle S \rangle$ is a subgroup of $A_5$, while the second is bad formulated: what is $[G: A/B]$? could you explain your notation? – Giorgio Mossa Feb 25 '12 at 16:58
Oh, OK. For Q2, try writing $[G:A]$ as $|G|/|A|$, etc. From there, try to get a $|G|$ on one side, and see what you get on the other. I will mention the obligatory "this works for infinite groups too", but the proof is different. – user641 Feb 25 '12 at 16:59
Hi Euden, and welcome to math.SE. I just wanted to say that I feel having the expression in question 2 on a separate line significantly improves readability, in addition to having all mathematical symbols rendered by LaTeX (including brackets), but if you insist you should feel free to change it back. However, the correct symbol in question 1 is $\subseteq$ (\subseteq), not $\supseteq$ (\supseteq). – Zev Chonoles Feb 25 '12 at 17:02
For Q1, your subgroup is actually contained in $A_4$. Does that help (what do you know of subgroups of order $6$ in $A_4$?) For Q2, please show some work so I can figure out where you're stuck, because the hint I gave should probably be enough. – user641 Feb 25 '12 at 17:57

For question 1: As has been noted, since $4$ is fixed by both generators of $G$, we can imagine that we are actually working in $S_4$ (acting on ${1,2,3,5}$). Moreover, since $3$-cycles are even, we are actually inside of $A_4$; $A_4$ has $12$ elements, and $S$ contains at least $5$ elements (namely, $\{1\}$, $(123)$, $(132)$, $(235)$, and $(253)$), it must either have $6$ elements or must be all of $A_4$ (by Lagrange's Theorem). However, $A_4$ is famous for being the smallest counterexample to the converse of Lagrange's Theorem: $A_4$ does not have subgroups of order $6$. Thus, $S$ must be all of $A_4$ (or rather, the even permutations in $S_5$ that fix $4$). If you want to explicitly produce the other seven elements explicitly in terms of the generators, we have (I compose my permutations right to left): \begin{align*} (123)(235) &= (12)(35)\\ (235)(123) &= (13)(25)\\ (123)(253) &= (125)\\ (253)(123) &= (153)\\ (132)(235) &= (135)\\ (235)(132) &= (152)\\ (152)(12)(35)(125) &= (15)(23) \end{align*} which gives all twelve elements of $A_4$.

For Question 2: Both the equality and the inequality hold for infinite groups.

Theorem. Let $K\leq H\leq G$ be groups; then $$[G:K]=[G:H][H:K]$$ in the sense of cardinalities.

Proof. Let $\{h_i\}_{i\in I}$ be a set of left coset representatives of $K$ in $H$; and let $\{g_j\}_{j\in J}$ be a set of left coset representatives for $H$ in $G$. We claim that $\{g_jh_i\}_{(i,j)\in I\times J}$ is a set of left coset representatives for $K$ in $G$.

Indeed, let $g\in G$. Then there exists $j\in J$ such that $g\in g_jH$; let $h\in H$ be such that $g=g_jh$. Then there exists $i\in I$ such that $h\in h_iK$. Therefore, there exists $k\in K$ such that $h=h_ik$. Hence, $g=g_jh_ik\in g_jh_i K$. That is: any element of $G$ is equivalent, modulo $K$ on the right, to some $g_jh_i$.

Now assume that $g_jh_iK = g_rh_sK$; we need to show that $j=r$ and $i=s$. Indeed, there exists $k\in K$ such that $g_jh_i = g_rh_sk$. Since $k\in K\subseteq H$, then $g_jh_i\in g_jH$, and $g_rh_sk\in g_rH$. Hence, $g_jH\cap g_rH\neq\varnothing$, hence $j=r$ (since the $g_j$ are a set of coset representatives). Thus, $g_j=g_r$, so $g_jh_iK=g_rh_sK$ implies $h_iK = h_sK$; since the $h_i$ form a set of coset representatives for $K$ in $H$, it follows that $i=s$, as desired. $\Box$

Theorem. Let $G$ be a group, and let $A$ and $B$ be subgroups. Then $$[A:A\cap B]\leq [G:B]$$ in the sense of cardinalities.

Proof. Let $\{a_i\}_{i\in I}$ be a set of coset representatives for $A\cap B$ in $A$. I claim that if $a_iB = a_jB$, then $i=j$. This will show that the number of left cosets of $B$ in $G$ is at least as large as $[A:A\cap B]$.

Assume that $a_iB=a_jB$. Then there exists $b\in B$ such that $a_i = a_jb$. Hence $b=a_ia_j^{-1}\in A\cap B$, so $a_i\in a_j(A\cap B)$; hence, $i=j$, as desired. $\Box$

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