1.3 Subgroups
Definition 1.3.1. A subset \(H\) of a group \(G\) is called a subgroup of \(G\)d if it is a group with respect
to operation in \(G\).
Note. If \(H\) is a subgroup of \(G\) and \(a\, , \, b\in H\), then \(ab\in H\).
- 1.
- \((\mathbb {Z}\, , \, +)\) is a subgroup of \(\, (\mathbb {R}\, , \, +)\).
- 2.
- With multiplication, \(\{1\, , \, -1\}\) is a subgroup of non-zero(identity is 1, but 0 has no multiplicative inverse, so we say non-zero) real numbers.
- 3.
- Every group is a subgroup of itself.
- 4.
- If \(e\) is the identity of a group, then \(\{e\}\) is a group of \(G\).
- 5.
- For all \(n,\, \, A_n\) is a subgroup of \(S_n\).
Notation:
If \(H\) is a subgroup of \(G\), then denote it as \(H\leq G\) or \(H<G\).
Lemma 1.3.3. Let \(G\) be a group with operation \(*\). Further, let \(H\) be a subgroup of \(G\).
- (a)
- If \(e'\) is the identity of \(H\) and \(e\) is the identity of \(G\), then \(e' = e\).
- (b)
- If \(a\in H\), then the inverse of \(a\) in \(H\) is the same as that of \(a\in G\).
Proof. Two cosets \(aH\) and \(bH\) are either equal or disjoint. Suppose they meet, say \(x\in aH\cap bH\), so \(x = ah_1 = bh_2\) for some \(h_1,h_2\in H\). Then \(a = bh_2h_1^{-1}\), and for any \(h\in H\), \[ah = b\left (h_2h_1^{-1}h\right ) \in bH ,\] so \(aH\subseteq bH\). Symmetry gives the reverse inclusion, hence \(aH = bH\).
Every element \(g\) lies in the coset \(gH\), since \(e\in H\), so the cosets cover \(G\). Being pairwise disjoint and covering, they partition \(G\). □
Remark. The map \(h\mapsto ah\) is a bijection from \(H\) to \(aH\), so every coset has exactly \(\left |H\right |\) elements. Combining that with the partition gives Lagrange’s theorem immediately: \(\left |G\right | = \left [G:H\right ]\left |H\right |\). This lemma is the whole content of that theorem, and the theorem is the first result in the subject that could not have been guessed from the definitions.
\(\{(1\,2)\, (1)\}\hspace {0.5cm}\) and \(\hspace {0.5cm} S_3 = {(1)\, , \, (1\, 2)\, , \, (1\, 3)\,, \cdots \cdots }\)
where \((1\,2)\) represents any element that is the multiplicative inverse of itself.
\(\bullet \, \{-1\, , \, 1\}\) with multiplication is a subgroup of \(S_3\).
Definition 1.3.6. Let \(G\) be a group and let \(H\) be a subset of \(G\). Then \(H\) is said to be a subgroup of \(G\) if and only if
- (a)
- \(H\) is non-empty
- (b)
- For all \(\, a,\, b\in H\, , \,\,\, ab\in H\).
- (c)
- If \(a\in H,\,\) then \(a^{-1}\in H\).
Proposition[one-step subgroup test]
A subset \(H\) of a group \(G\) is a subgroup of \(G\) if and only if \(H\neq \emptyset \) and whenever \(x,\, y\in H\, , \, \, xy^{-1}\in H\).
Proof. If \(H\) is a subgroup, then \(H \neq \emptyset \), since \(e\in H\). If \(x,\, y\in H\), then \(y^{-1}\in H\) by (c) in definition so that by (b) \(xy^{-1}\in H\).
Conversely, suppose \(H\) is a subset of \(G\) such that \(H\neq \emptyset \) and whenever \(x,\, y\in H\) then \(xy^{-1}\in H\). Now since \(H\neq \emptyset \), it must contain
an element say \(h\). Take \(x = y = h\), we see that \(e = hh^{-1}\in H\). If \(y\in H\), set \(x= e\), we have \(y^{-1} = ey^{-1}\in H\). Finally, we know that \((y^{-1})^{-1} = y\). Hence, \(x\,, \, y\in H\), then \(y^{-1}\in H = x(y^{-1})^{-1}\in H\).
Therefore \(H\) is a subgroup.
□
- 1.
- Let \(G\) an Abelian group with identity \(e\). Then \(H = \{x\in G:\, x^2 = e\}\) is a subgroup of \(G\). \(\big [\)Torston elts\(\big ]\).
Note. Clearly, \(H\neq \emptyset ,\,\, e\in H\) since \(e^2 = e\). For each \(a, \, b\in H\) we have by definition, \(a^2 = e\, ,\)
\(\, b^2 = e\). We need to show that \(ab^{-1}\in H\) i.e \((ab^{-1})^2 = e\). Now\begin {align*} (ab^{-1})^2 & = (ab^{-1})(ab^{-1}) \\ & = aab^{-1}b^{-1}\hspace {0.5cm}(\text {Since}\hspace {0.2cm}G\hspace {0.2cm}\text {is Abelian})\\ & = a^2 (b^{-1})^2 = a^2(b^2)^{-1}\\ & = e(e)^{-1} = e^2 = e \end {align*}
Therefore \(\, ab^{-1}\in H\), thus \(\, H<G\).
- 2.
- Let \(G = \mathbb {Z}_8\) (integers of modulo 8). Then \(H = \mathbb {Z}_{8/2}\) is a subgroup of \(G\).
\[\mathbb {Z}_8 = \{0,\, 1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7\}\hspace {0.4cm}\text {and}\hspace {0.4cm} \mathbb {Z}_{8/2} = \{0,\, 2,\, 4,\, 6\}\]
\(Z_n \implies \) addition.
Note. \(H\neq \emptyset \)
Now, for \(x\, y\in H\) we need to show that \(\, xy^{-1}\in H\)\(y\)\(+\) 0 6 4 2 \(x\)0 0 6 4 2 2 2 0 6 4 4 4 2 0 4 6 6 4 2 0 \(=xy^{-1} \in H\) since base 8\(0^{-1} = 0\) since \(0 + 0 = 0\) \(2^{-1} = 6\) \(2 + 6 = 8 = 0\) \(4^{-1} = 4\) \(4 + 4 = 8 = 0\) \(6^{-1} = 2\) \( 6 + 2 = 8 = 0\) \(xy^{-1} = x-y\) - 3.
- Show that the set \(H = \{(1)\, , \, (1\, 2\, 3)\}\) is not a subgroup of \(S_3\).
Note. Let \(x = y = (1\, 2\, 3)\)
\((1\, 2\, 3)(3\, 2\, 1) = \begin {pmatrix} 1 & 2 & 3\\ 2 & 3 & 1\\ \end {pmatrix}\begin {pmatrix} 1 & 2 & 3\\ 3 & 1 & 2\\ \end {pmatrix}=\begin {pmatrix} 1 & 2 & 3\\ 1 & 2 & 3\\ \end {pmatrix}\)
Let \(x = (1)\) and \(y = (1\, 2\, 3)\,\,\, y^{-1} = (3\, 2\, 1)\implies xy^{-1} = (1)(3\, 2\, 1)\not \in H\). Hence
Exercise 1.3.8. Show that \(\, A_n < S_n \, \, \forall n\).
Solution. (By nemwine!!!)
Let \(S_n = \{a_1\, , \, a_2\, \cdots \cdots \, ,\, a_n\}\). Then \(A_n = \{a_i:\, i = 1, \, 2, \, \cdots \cdots \}\) where \(a_i\) is even and \(a_i\in S_n\). Now, since \(\begin {vmatrix} S_n\\ \end {vmatrix} = n!\) we have that \(\begin {vmatrix} A_n\\ \end {vmatrix} = \dfrac {n!}{2}\), since the \(\begin {vmatrix} A_n\\ \end {vmatrix}\) divides \(S_n\) into two sets of
odd and even permutations. This implies if \(\begin {vmatrix} S_n\\ \end {vmatrix} = n!\) and \(\begin {vmatrix} A_n\\ \end {vmatrix} = k!\) then \(k! = \dfrac {n!}{2}\, \implies k! < n! \implies \begin {vmatrix} A_n\\ \end {vmatrix} < \begin {vmatrix} S_n\\ \end {vmatrix} \implies A_n < S_n\,\,\forall n\in N\).
Theorem 1.3.9. Let \(G\) be a group and \(H\) and \(K\) be subgroups of \(G\). Then \(H\cap K\) is a subgroup of \(G\).
Proof. \(H\cap K\neq \emptyset \) since \(e\in H\) and \(e\in K\implies e\in H\cap K\). Now, let \(h\in H\cap K\), then \(h\in H\) and \(h\in K\). But \(H\) is a subgroup, then \(h^{-1}\in H\) and \(h^{-1}\in K\). Thus, \(h^{-1}\in H\cap K\). Finally, let \(h,\,h'\in H\cap K\),
then \(h\in H\) and \(h'\in H\). Now \(H\) is a subgroup so \(hh'\in H\) similarly \(h\in K\) and \(h'\in K\), \(K\) is a subgroup, \(hh'\in K\). Thus \(hh'\in H\cap K\). Thus \(H\cap K\) is a subgroup.
□
Note. Given subgroup \(H\, , \, K\) in \(G\). Then union \(H\cup K\) is not always a subgroup of \(G\).
\(\mathbb {Z}_{6/2}\cup Z_{6/3}\, \) is not a subgroup of \(\mathbb {Z}_6\) \[\mathbb {Z}_6 = \{0,\, 1,\, 2,\, 3,\, 4, \, 5\}\hspace {0.5cm} \mathbb {Z}_{6/2}= \{0,\, 2,\, 4\}\hspace {0.5cm} \mathbb {Z}_{6/3} = \{0,\, 3\}\] \[\implies \, \mathbb {Z}_{6/2}\cup \mathbb {Z}_{6/3} \, \cong \, \mathbb {Z}_3\cup \mathbb {Z}_2 = \{0,\, 2, \, 3,\, 4\}\]
| \(+\) | \(0^{-1} = 0\) | \(2^{-1} = 4\) | \(3^{-1} = 3\) | \(4^{-1} = 2\) |
| 0 | 0 | 4 | 3 | 2 |
| 2 | 2 | 0 | 5 | 4 |
| 3 | 3 | 1 | 0 | 5 |
| 4 | 4 | 2 | 1 | 0 |
\(5\not \in \{0,\, 2,\, 3,\,4\} = \mathbb {Z}_3\cup \mathbb {Z}_2.\hspace {0.3cm} \therefore \, \mathbb {Z}_3\cup \mathbb {Z}_2\,\) is not a subgroup.
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