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If a subset \(H\) of a group \(G\) is closed under the binary operation of \(G\) and if \(H\) with the induced operation from \(G\) is itself a group, then \(H\) is a subgroup of \(G\).
Denoted by \(H \leq G\) or \(G \geq H\).
Example
Under addition, we have \(\mathbb{Z \leq Q \leq R \leq C}\).
If \(G\) is a group, then the subgroup consisting of \(G\) itself is the improper subgroup of \(G\). All other subgroups are proper subgroups of \(G\).
The subgroup \(\{e\}\) is the trivial subgroup of \(G\). All other subgroups are nontrivial.
Let \(H\) be a subgroup of \(G\). Then the identity \(e_H\) of \(H\) is equal to the identity \(e_G\) of \(G\). Moreover, for all \(a \in H\), the inverse of \(a \in H\) is equal to the inverse of \(a \in G\).
Proof
By the definitions of identity and inverse.
A subset \(H\) of a group \(G\) is a subgroup of \(G\) if and only if
- \(H\) is closed under the binary operation of \(G\),
- \(H\) contains \(e_G\) .
- for all \(a \in H\), the inverse \(a^{-1}\) is also in \(H\).
其實 1. 和 3. imply 2.，但 2. 的必要性在於防止 \(H\) 是空集合，不過空集合作為 subgroup 沒什麼意義（是嗎？）。
\(H\) is closed，但沒有說 for \(g_1,g_2 \not \in H,\ g_1g_2 \in H\) 是錯的！（外部可闖入；可用 \(\mathbb{Z}_{12}\) 找例子。）
Q1:
Let \(G = \mathbb{Z}\times\mathbb{Z}, H = \langle(2,1),(1,2)\rangle,\) and \(N = \langle(3, 0), (0, 3) \rangle\). Show that \(N \le H\).
Solution
It is obivous that \(H\le G\) and \(N\le G\) since they are defined as groups. Thus it suffices to show that \(N\) is a subset of \(H\): For all \(h \in H\), \(h = (2a+b, a+2b)\), where \(a,b \in \mathbb{Z}\); for all \(g\in N\), \(g=(3c, 3d)\), where \(c,d\in \mathbb{Z}\). To show that for all \(g\in N\), \(g\) is also in \(H\), we have
\[2a+b = 3c, a+2b = 3d,\]which means that \(a=2c-d\) and \(b=2d-c\) and they are both integers. Thus \(N\subseteq H\) and \(N\le H\). ◼
Q2:
Let \(g\) be an element if a group \(G\). Show that the following is a subgroup:
\[C_G(g) = \{x\in G\vert gx=xg\}.\]Solution
closedness
For all \(x,y \in C_G(g)\), \(gx = xg\) and \(gy = yg\). Then \(gxy = xgy = xyg\), which means that \(xy \in C_G(g)\).
identity
Since \(eg = e = ge\), \(e\in C_G(g)\).
inverse
For all \(x \in C_G(g)\), \(gx = xg\). Multiply both sides of the equation by \(x^{-1}\) from both left and right, we obtain \(x^{-1}g = gx^{-1}\). We conclude that \(x^{-1}\in C_G(g)\).
Combing the above results, \(C_G(g)\) is a subgroup. ◼