Let (G,βˆ—)(G, *) be a group and HβŠ‚GH \subset G. HH is called a subgroup of GG (usually denoted by H≀GH \leq G ) if HH also forms a group with respect to the restriction of the same binary operation βˆ—*, i.e.,

H1 for all a,b∈Hβ‡’aβˆ—b∈Ha, b \in H \Rightarrow a * b \in H (HH is closed with respect to βˆ—).*) .

H2 HH together with the induced binary operation.

βˆ—:HΓ—Hβ†’H,(a,b)↦aβˆ—b*: H \times H \rightarrow H,(a, b) \mapsto a * b

forms a group, meaning that the restriction of βˆ—* to HΓ—HH \times H is a binary group operation on H\mathrm{H}.

In order to form a subgroup of any group G,HG, H has to fulfill the following minimal requirements:

Lemma 1: Subgroup criteria

Let (G,βˆ—)(G, *) be a group, HβŠ‚GH \subset G with Hβ‰ βˆ….H \neq \varnothing . Then, the following statements are equivalent:

  1. Hβ©½GH \leqslant G

  2. a,b∈Hβ‡’aβˆ—b∈Ha, b \in H \Rightarrow a * b \in H and aβˆ’1∈Ha^{-1} \in H

  3. a,b∈Hβ‡’aβˆ—bβˆ’1∈Ha, b \in H \Rightarrow a * b^{-1} \in H


  • (1) β‡’\Rightarrow (2): Let Hβ©½GH \leqslant G. According to the above definition, there are a,b∈Ha, b \in H and also aβˆ—b∈Ha * b \in H. Since HH is a group, there exists an identity element eβ€²e^{\prime}. Because eβ€²βˆ—eβ€²=eβ€²e^{\prime} * e^{\prime}=e^{\prime}, as the concept discussed here :Corollary_2_6_1 yields eβ€²=ee^{\prime}=e. Furthermore, since HH is a group, there exists an inverse aβ€²βˆˆHa^{\prime} \in H for every a∈Ha \in H such that aβ€²βˆ—a=ea^{\prime} * a=e. Therefore, aβˆ’1=aβ€²βˆˆHa^{-1}=a^{\prime} \in H.

  • (2) β‡’\Rightarrow (3): According to statement (2), if a,b∈Ha, b \in H, then a∈Ha \in H and also bβˆ’1∈Hb^{-1} \in H, therefore (2) yields aβˆ—bβˆ’1∈Ha * b^{-1} \in H.

  • (3) β‡’\Rightarrow (1): For a,b∈Ha, b \in H, (3) yields bβˆ—bβˆ’1∈Hb * b^{-1} \in H, and thus e=bβˆ—bβˆ’1∈He=b * b^{-1} \in H. It follows that bβˆ’1=eβˆ—bβˆ’1b^{-1}=e * b^{-1} for all b∈Hb \in H, and therefore aβˆ—b=aβˆ—(bβˆ’1)βˆ’1a * b=a *\left(b^{-1}\right)^{-1}. Altogether, a,b∈Hβ‡’aβˆ—b∈Ha, b \in H \Rightarrow a * b \in H, which yields H≀GH \leq G according to the definition given above (H1).

These minimal demands are defined in this lemma together with this lemma :Lemma_2_6_2 yield very natural conditions for an easy verifying of HH to be a subgroup of GG.

Theorem 1: subgroup verification

Hβ©½GH \leqslant G if, and only if, the following conditions are satisfied:

  1. HH is closed with respect to the binary operator βˆ—*, i.e., aβˆ—b∈Ha * b \in H for every a,b∈Ha, b \in H.

  2. H\mathrm{H} contains the identity element eGe_{G} of GG.

  3. For each a∈H,Ha \in H, H also contains its inverse element, i.e., if a,aβˆ’1∈Ga, a^{-1} \in G and a∈Ha \in H, then also aβˆ’1∈Ha^{-1} \in H.

Note: For any group G=(G,βˆ—)G=(G, *) with identity eG,{eG}e_{G},\left\{e_{G}\right\} and GG itself are always subgroups of GG, whereas {eG}\left\{e_{G}\right\} is called the trivial subgroup. Any other subgroup is said to be a proper subgroup.


(Z,+)(\mathbb{Z},+) is a subgroup of (Q,+)(\mathbb{Q},+) and (R,+)(\mathbb{R},+). It’s easy to see that the set of integers Z\mathbb{Z} is closed with respect to ++, meaning that a+b∈Za+b \in \mathbb{Z} for every a,b∈Za, b \in \mathbb{Z}. Furthermore, the identity element 0 of Q\mathbb{Q} and R\mathbb{R} is also contained in Z\mathbb{Z}. Additionally, Z\mathbb{Z} contains the inverse βˆ’a-a for all a∈Za \in \mathbb{Z}.


(Zn,+)\left(\mathbb{Z}_{n},+\right) is a subgroup of (Z,+)(\mathbb{Z},+).

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